Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 12 Algebraic Identities

Exercise 12(A)

Question 1. Use direct method to evaluate the following products :
(i) \( (x + 8)(x + 3) \)
(ii) \( (y + 5)(y - 3) \)
(iii) \( (a - 8)(a + 2) \)
(iv) \( (b - 3)(b - 5) \)
(v) \( (3x - 2y)(2x + y) \)
(vi) \( (5a + 16)(3a - 7) \)
(vii) \( (8 - b) (3 + b) \)
Answer:
(i) \( (x + 8)(x + 3) = (x \times x) + (x \times 3) + (8 \times x) + (8 \times 3) \)
\( = x^2 + 3x + 8x + 24 \)
\( = x^2 + 11x + 24 \)

(ii) \( (y + 5)(y - 3) = (y \times y) + (y \times -3) + (5 \times y) + (5 \times -3) \)
\( = y^2 + (-3y) + (5y) - 15 \)
\( = y^2 - 3y + 5y - 15 \)
\( = y^2 + 2y - 15 \)

(iii) \( (a - 8)(a + 2) = (a \times a) + (a \times 2) + (-8) \times a + (-8)(2) \)
\( = a^2 + 2a - 8a - 16 \)
\( = a^2 - 6a - 16 \)

(iv) \( (b - 3)(b - 5) = (b \times b) + (b \times -5) + (-3) \times b + (-3)(-5) \)
\( = b^2 - 5b - 3b + 15 \)
\( = b^2 - 8b + 15 \)

(v) \( (3x - 2y)(2x + y) = (3x \times 2x) + (3x \times y) + (-2y \times 2x) + (-2y \times y) \)
\( = 6x^2 + 3xy - 4xy - 2y^2 \)
\( = 6x^2 - xy - 2y^2 \)

(vi) \( (5a + 16)(3a - 7) = (5a \times 3a) + (5a \times -7) + (16 \times 3a) + (16 \times -7) \)
\( = 15a^2 + (-35a) + 48a + (-112) \)
\( = 15a^2 - 35a + 48a - 112 \)
\( = 15a^2 + 13a - 112 \)

(vii) \( (8 - b)(3 + b) = (8 \times 3) + (8 \times b) + (-b \times 3) + (-b \times b) \)
\( = 24 + 8b - 3b - b^2 \)
\( = 24 + 5b - b^2 \)
In simple words: Multiply each term in the first bracket by each term in the second bracket individually, then simplify by adding or subtracting the terms that are alike.

Exam Tip: Be very careful with signs when distributing negative terms, as multiplying two negatives always yields a positive result.

Question 2. Use direct method to evaluate :
(i) \( (x + 1)(x - 1) \)
(ii) \( (2 + a)(2 - a) \)
(iii) \( (3b - 1)(3b + 1) \)
(iv) \( (4 + 5x)(4 - 5x) \)
(v) \( (2a + 3)(2a - 3) \)
(vi) \( (xy + 4)(xy - 4) \)
(vii) \( (ab + x^2)(ab - x^2) \)
(viii) \( (3x^2 + 5y^2)(3x^2 - 5y^2) \)
(ix) \( \left(z - \frac{2}{3}\right)\left(z + \frac{2}{3}\right) \)
(x) \( \left(\frac{3}{5}a + \frac{1}{2}\right)\left(\frac{3}{5}a - \frac{1}{2}\right) \)
(xi) \( (0.5 - 2a)(0.5 + 2a) \)
(xii) \( \left(\frac{a}{2} - \frac{b}{3}\right)\left(\frac{a}{2} + \frac{b}{3}\right) \)
Answer:
Note: Using the algebraic identity \( (a + b)(a - b) = a^2 - b^2 \):
(i) \( (x + 1)(x - 1) = (x)^2 - (1)^2 = x^2 - 1 \)

(ii) \( (2 + a)(2 - a) = (2)^2 - (a)^2 = 4 - a^2 \)

(iii) \( (3b - 1)(3b + 1) = (3b)^2 - (1)^2 = 9b^2 - 1 \)

(iv) \( (4 + 5x)(4 - 5x) = (4)^2 - (5x)^2 = 16 - 25x^2 \)

(v) \( (2a + 3)(2a - 3) = (2a)^2 - (3)^2 = 4a^2 - 9 \)

(vi) \( (xy + 4)(xy - 4) = (xy)^2 - (4)^2 = x^2y^2 - 16 \)

(vii) \( (ab + x^2)(ab - x^2) = (ab)^2 - (x^2)^2 = a^2b^2 - x^4 \)

(viii) \( (3x^2 + 5y^2)(3x^2 - 5y^2) = (3x^2)^2 - (5y^2)^2 = 9x^4 - 25y^4 \)

(ix) \( \left(z - \frac{2}{3}\right)\left(z + \frac{2}{3}\right) = (z)^2 - \left(\frac{2}{3}\right)^2 = z^2 - \frac{4}{9} \)

(x) \( \left(\frac{3}{5}a + \frac{1}{2}\right)\left(\frac{3}{5}a - \frac{1}{2}\right) = \left(\frac{3}{5}a\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{9}{25}a^2 - \frac{1}{4} \)

(xi) \( (0.5 - 2a)(0.5 + 2a) = (0.5)^2 - (2a)^2 = 0.25 - 4a^2 \)

(xii) \( \left(\frac{a}{2} - \frac{b}{3}\right)\left(\frac{a}{2} + \frac{b}{3}\right) = \left(\frac{a}{2}\right)^2 - \left(\frac{b}{3}\right)^2 = \frac{a^2}{4} - \frac{b^2}{9} \)
In simple words: When multiplying a sum and a difference of the exact same two terms, the product is always the square of the first term minus the square of the second term.

Exam Tip: Remember to square both the coefficient and the variable when evaluating terms like \( (3b)^2 \), which becomes \( 9b^2 \), not \( 3b^2 \).

Question 3. Evaluate :
(i) \( (a + 1)(a - 1)(a^2 + 1) \)
(ii) \( (a + b)(a - b)(a^2 + b^2) \)
(iii) \( (2a - b)(2a + b)(4a^2 + b^2) \)
(iv) \( (3 - 2x)(3 + 2x)(9 + 4x^2) \)
(v) \( (3x - 4y)(3x + 4y)(9x^2 + 16y^2) \)
Answer:
(i) \( (a + 1)(a - 1)(a^2 + 1) = [(a)^2 - (1)^2](a^2 + 1) \)
\( = (a^2 - 1)(a^2 + 1) \)
\( = (a^2)^2 - (1)^2 \)
\( = a^4 - 1 \)

(ii) \( (a + b)(a - b)(a^2 + b^2) = (a^2 - b^2)(a^2 + b^2) \)
\( = (a^2)^2 - (b^2)^2 \)
\( = a^4 - b^4 \)

(iii) \( (2a - b)(2a + b)(4a^2 + b^2) = [(2a)^2 - (b)^2](4a^2 + b^2) \)
\( = (4a^2 - b^2)(4a^2 + b^2) \)
\( = (4a^2)^2 - (b^2)^2 \)
\( = 16a^4 - b^4 \)

(iv) \( (3 - 2x)(3 + 2x)(9 + 4x^2) = [(3)^2 - (2x)^2](9 + 4x^2) \)
\( = (9 - 4x^2)(9 + 4x^2) \)
\( = (9)^2 - (4x^2)^2 \)
\( = 81 - 16x^4 \)

(v) \( (3x - 4y)(3x + 4y)(9x^2 + 16y^2) = [(3x)^2 - (4y)^2](9x^2 + 16y^2) \)
\( = (9x^2 - 16y^2)(9x^2 + 16y^2) \)
\( = (9x^2)^2 - (16y^2)^2 \)
\( = 81x^4 - 256y^4 \)
In simple words: Apply the difference of squares identity to the first two brackets, and then apply the identity a second time to the resulting bracket and the last bracket.

Exam Tip: This is a chain of identities. Always evaluate from left to right, combining the first pair of factors first to reveal the next identity.

Question 4. Use the product (a + b)(a - b) = a^2 - b^2 to evaluate:
(i) \( 21 \times 19 \)
(ii) \( 33 \times 27 \)
(iii) \( 103 \times 97 \)
(iv) \( 9.8 \times 10.2 \)
(v) \( 7.7 \times 8.3 \)
(vi) \( 4.6 \times 5.4 \)
Answer:
(i) \( 21 \times 19 = (20 + 1)(20 - 1) \)
\( = (20)^2 - (1)^2 = 400 - 1 = 399 \)

(ii) \( 33 \times 27 = (30 + 3)(30 - 3) \)
\( = (30)^2 - (3)^2 = 900 - 9 = 891 \)

(iii) \( 103 \times 97 = (100 + 3)(100 - 3) \)
\( = (100)^2 - (3)^2 = 10000 - 9 = 9991 \)

(iv) \( 9.8 \times 10.2 = (10 - 0.2)(10 + 0.2) \)
\( = (10)^2 - (0.2)^2 = 100 - 0.04 = 99.96 \)

(v) \( 7.7 \times 8.3 = (8 - 0.3)(8 + 0.3) \)
\( = (8)^2 - (0.3)^2 = 64 - 0.09 = 63.91 \)

(vi) \( 4.6 \times 5.4 = (5 - 0.4)(5 + 0.4) \)
\( = (5)^2 - (0.4)^2 = 25 - 0.16 = 24.84 \)
In simple words: Express the given numbers as a sum and difference of a round number and a small value, then use the shortcut formula to find the product without multiplying long numbers.

Exam Tip: Find the number exactly mid-way between the two numbers to act as your term \( a \), and the offset from that midpoint will be your term \( b \).

Question 5. Evaluate :
(i) \( (6 - xy)(6 + xy) \)
(ii) \( \left(7x + \frac{2}{3}y\right)\left(7x - \frac{2}{3}y\right) \)
(iii) \( \left(\frac{a}{2b} + \frac{2b}{a}\right)\left(\frac{a}{2b} - \frac{2b}{a}\right) \)
(iv) \( \left(3x - \frac{1}{2y}\right)\left(3x + \frac{1}{2y}\right) \)
(v) \( (2a + 3)(2a - 3)(4a^2 + 9) \)
(vi) \( (a + bc)(a - bc)(a^2 + b^2c^2) \)
(vii) \( (5x + 8y)(3x + 5y) \)
(viii) \( (7x + 15y)(5x - 4y) \)
(ix) \( (2a - 3b)(3a + 4b) \)
(x) \( (9a - 7b)(3a - b) \)
Answer:
(i) \( (6 - xy)(6 + xy) = 6(6 + xy) - xy(6 + xy) \)
\( = 36 + 6xy - 6xy - x^2y^2 = 36 - x^2y^2 \)

(ii) \( \left(7x + \frac{2}{3}y\right)\left(7x - \frac{2}{3}y\right) = 7x\left(7x - \frac{2}{3}y\right) + \frac{2}{3}y\left(7x - \frac{2}{3}y\right) \)
\( = 49x^2 - \frac{14}{3}xy + \frac{14}{3}xy - \frac{4}{9}y^2 = 49x^2 - \frac{4}{9}y^2 \)

(iii) \( \left(\frac{a}{2b} + \frac{2b}{a}\right)\left(\frac{a}{2b} - \frac{2b}{a}\right) = \frac{a}{2b}\left(\frac{a}{2b} - \frac{2b}{a}\right) + \frac{2b}{a}\left(\frac{a}{2b} - \frac{2b}{a}\right) \)
\( = \frac{a^2}{4b^2} - 1 + 1 - \frac{4b^2}{a^2} = \frac{a^2}{4b^2} - \frac{4b^2}{a^2} \)

(iv) \( \left(3x - \frac{1}{2y}\right)\left(3x + \frac{1}{2y}\right) = 3x\left(3x + \frac{1}{2y}\right) - \frac{1}{2y}\left(3x + \frac{1}{2y}\right) \)
\( = 9x^2 + \frac{3x}{2y} - \frac{3x}{2y} - \frac{1}{4y^2} = 9x^2 - \frac{1}{4y^2} \)

(v) \( (2a + 3)(2a - 3)(4a^2 + 9) = [(2a)^2 - (3)^2](4a^2 + 9) \)
\( = (4a^2 - 9)(4a^2 + 9) \)
\( = (4a^2)^2 - (9)^2 = 16a^4 - 81 \)

(vi) \( (a + bc)(a - bc)(a^2 + b^2c^2) = [a^2 - (bc)^2](a^2 + b^2c^2) \)
\( = (a^2 - b^2c^2)(a^2 + b^2c^2) \)
\( = (a^2)^2 - (b^2c^2)^2 = a^4 - b^4c^4 \)

(vii) \( (5x + 8y)(3x + 5y) = 5x(3x + 5y) + 8y(3x + 5y) \)
\( = 15x^2 + 25xy + 24xy + 40y^2 \)
\( = 15x^2 + 49xy + 40y^2 \)

(viii) \( (7x + 15y)(5x - 4y) = 7x(5x - 4y) + 15y(5x - 4y) \)
\( = 35x^2 - 28xy + 75xy - 60y^2 \)
\( = 35x^2 + 47xy - 60y^2 \)

(ix) \( (2a - 3b)(3a + 4b) = 2a(3a + 4b) - 3b(3a + 4b) \)
\( = 6a^2 + 8ab - 9ab - 12b^2 \)
\( = 6a^2 - ab - 12b^2 \)

(x) \( (9a - 7b)(3a - b) = 9a(3a - b) - 7b(3a - b) \)
\( = 27a^2 - 9ab - 21ab + 7b^2 \)
\( = 27a^2 - 30ab + 7b^2 \)
In simple words: When the brackets match the \( (u+v)(u-v) \) pattern, you can use the difference of squares directly. For general cases with different numbers, multiply the expressions using the distributive law.

Exam Tip: Be careful with fraction variables in denominators. When multiplying \( \frac{a}{2b} \times \frac{2b}{a} \), terms cancel out completely leaving a constant \( 1 \).

Exercise 12(B)

Question 1. Expand :
(i) \( (2a + b)^2 \)
(ii) \( (a - 2b)^2 \)
(iii) \( \left(a + \frac{1}{2a}\right)^2 \)
(iv) \( \left(2a - \frac{1}{a}\right)^2 \)
(v) \( (a + b - c)^2 \)
(vi) \( (a - b + c)^2 \)
(vii) \( \left(3x + \frac{1}{3x}\right)^2 \)
(viii) \( \left(2x - \frac{1}{2x}\right)^2 \)
Answer:
(i) \( (2a + b)^2 = (2a)^2 + (b)^2 + 2 \times 2a \times b \)
\( = 4a^2 + b^2 + 4ab \)

(ii) \( (a - 2b)^2 = (a)^2 + (2b)^2 - 2 \times a \times 2b \)
\( = a^2 + 4b^2 - 4ab \)

(iii) \( \left(a + \frac{1}{2a}\right)^2 = (a)^2 + \left(\frac{1}{2a}\right)^2 + 2 \times a \times \frac{1}{2a} \)
\( = a^2 + \frac{1}{4a^2} + \frac{2a}{2a} \)
\( = a^2 + \frac{1}{4a^2} + 1 \)

(iv) \( \left(2a - \frac{1}{a}\right)^2 = (2a)^2 + \left(\frac{1}{a}\right)^2 - 2 \times 2a \times \frac{1}{a} \)
\( = 4a^2 + \frac{1}{a^2} - 4 \)

(v) \( (a + b - c)^2 = (a)^2 + (b)^2 + (-c)^2 + 2 \times a \times b + 2 \times b \times (-c) + 2 \times (-c) \times a \)
\( = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca \)

(vi) \( (a - b + c)^2 = (a)^2 + (-b)^2 + (c)^2 + 2 \times a \times (-b) + 2 \times (-b) \times c + 2 \times c \times a \)
\( = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca \)

(vii) \( \left(3x + \frac{1}{3x}\right)^2 = (3x)^2 + \left(\frac{1}{3x}\right)^2 + 2 \times 3x \times \frac{1}{3x} \)
\( = 9x^2 + \frac{1}{9x^2} + 2 \)

(viii) \( \left(2x - \frac{1}{2x}\right)^2 = (2x)^2 + \left(\frac{1}{2x}\right)^2 - 2 \times 2x \times \frac{1}{2x} \)
\( = 4x^2 + \frac{1}{4x^2} - 2 \)
In simple words: Use the standard squaring formulas \( (x+y)^2 = x^2 + y^2 + 2xy \) and \( (x-y)^2 = x^2 + y^2 - 2xy \). For expressions with three terms, square each term and then add twice the products of all possible pairs, keeping their respective negative signs.

Exam Tip: When evaluating reciprocal terms like \( 2 \times a \times \frac{1}{2a} \), the variables cancel out leaving just a pure numeric value.

Question 2. Find the square of :
(i) \( x + 3y \)
(ii) \( 2x - 5y \)
(iii) \( a + \frac{1}{5a} \)
(iv) \( 2a - \frac{1}{a} \)
(v) \( x - 2y + 1 \)
(vi) \( 3a - 2b - 5c \)
(vii) \( 2x + \frac{1}{x} + 1 \)
(viii) \( 5 - x + \frac{2}{x} \)
(ix) \( 2x - 3y + z \)
(x) \( x + \frac{1}{x} - 1 \)
Answer:
(i) \( (x + 3y)^2 = (x)^2 + (3y)^2 + 2 \times x \times 3y \)
\( = x^2 + 9y^2 + 6xy \)

(ii) \( (2x - 5y)^2 = (2x)^2 + (5y)^2 - 2 \times 2x \times 5y \)
\( = 4x^2 + 25y^2 - 20xy \)

(iii) \( \left(a + \frac{1}{5a}\right)^2 = (a)^2 + \left(\frac{1}{5a}\right)^2 + 2 \times a \times \frac{1}{5a} \)
\( = a^2 + \frac{1}{25a^2} + \frac{2}{5} \)

(iv) \( \left(2a - \frac{1}{a}\right)^2 = (2a)^2 + \left(\frac{1}{a}\right)^2 - 2 \times 2a \times \frac{1}{a} \)
\( = 4a^2 + \frac{1}{a^2} - 4 \)

(v) \( (x - 2y + 1)^2 = (x)^2 + (-2y)^2 + (1)^2 + 2 \times x \times (-2y) + 2 \times (-2y) \times 1 + 2 \times 1 \times x \)
\( = x^2 + 4y^2 + 1 - 4xy - 4y + 2x \)

(vi) \( (3a - 2b - 5c)^2 = (3a)^2 + (-2b)^2 + (-5c)^2 + 2 \times 3a \times (-2b) + 2 \times (-2b) \times (-5c) + 2 \times (-5c) \times 3a \)
\( = 9a^2 + 4b^2 + 25c^2 - 12ab + 20bc - 30ca \)

(vii) \( \left(2x + \frac{1}{x} + 1\right)^2 = (2x)^2 + \left(\frac{1}{x}\right)^2 + (1)^2 + 2 \times 2x \times \frac{1}{x} + 2 \times \frac{1}{x} \times 1 + 2 \times 1 \times 2x \)
\( = 4x^2 + \frac{1}{x^2} + 1 + 4 + \frac{2}{x} + 4x \)
\( = 4x^2 + \frac{1}{x^2} + 5 + \frac{2}{x} + 4x \)

(viii) \( \left(5 - x + \frac{2}{x}\right)^2 = (5)^2 + (-x)^2 + \left(\frac{2}{x}\right)^2 + 2 \times 5 \times (-x) + 2 \times (-x) \times \frac{2}{x} + 2 \times \frac{2}{x} \times 5 \)
\( = 25 + x^2 + \frac{4}{x^2} - 10x - 4 + \frac{20}{x} \)
\( = 21 + x^2 + \frac{4}{x^2} - 10x + \frac{20}{x} \)

(ix) \( (2x - 3y + z)^2 = (2x)^2 + (-3y)^2 + (z)^2 + 2 \times 2x \times (-3y) + 2 \times (-3y) \times z + 2 \times z \times 2x \)
\( = 4x^2 + 9y^2 + z^2 - 12xy - 6yz + 4zx \)

(x) \( \left(x + \frac{1}{x} - 1\right)^2 = (x)^2 + \left(\frac{1}{x}\right)^2 + (-1)^2 + 2 \times x \times \frac{1}{x} + 2 \times \frac{1}{x} \times (-1) + 2 \times (-1) \times x \)
\( = x^2 + \frac{1}{x^2} + 1 + 2 - \frac{2}{x} - 2x \)
\( = x^2 + \frac{1}{x^2} + 3 - \frac{2}{x} - 2x \)
In simple words: To square a binomial or trinomial, write it raised to the power of 2 and expand it using the respective identity. Watch out for negative terms.

Exam Tip: For three-term expressions, group constants together or simplify the constants first, as seen when \( 2 \times x \times \frac{1}{x} = 2 \) merges with \( (1)^2 = 1 \) to form \( 3 \).

Question 3. Evaluate using expansion of (a + b)^2 or (a - b)^2:
(i) \( (208)^2 \)
(ii) \( (92)^2 \)
(iii) \( (415)^2 \)
(iv) \( (188)^2 \)
(v) \( (9.4)^2 \)
(vi) \( (20.7)^2 \)
Answer:
(i) \( (208)^2 = (200 + 8)^2 \)
\( = (200)^2 + (8)^2 + 2(200)(8) = 40000 + 64 + 3200 \)
\( = 43264 \)

(ii) \( (92)^2 = (100 - 8)^2 = (100)^2 + (8)^2 - 2(100)(8) \)
\( = 10000 + 64 - 1600 = 10064 - 1600 = 8464 \)

(iii) \( (415)^2 = (400 + 15)^2 \)
\( = (400)^2 + (15)^2 + 2(400)(15) = 160000 + 225 + 12000 \)
\( = 172225 \)

(iv) \( (188)^2 = (200 - 12)^2 \)
\( = (200)^2 + (12)^2 - 2(200)(12) = 40000 + 144 - 4800 \)
\( = 40144 - 4800 = 35344 \)

(v) \( (9.4)^2 = (10 - 0.6)^2 \)
\( = (10)^2 + (0.6)^2 - 2(10)(0.6) = 100 + 0.36 - 12 \)
\( = 88 + 0.36 = 88.36 \)

(vi) \( (20.7)^2 = (20 + 0.7)^2 = (20)^2 + (0.7)^2 + 2(20)(0.7) \)
\( = 400 + 0.49 + 28 = 428 + 0.49 = 428.49 \)
In simple words: Break each number into an easy base number (like 10, 100, or 200) plus or minus a smaller value, then use squaring identities for easy calculations.

Exam Tip: Try to choose a base number ending in 0 (like 10, 100, or 200) to make the calculations quick and less prone to multiplication errors.

Question 4. Expand :
(i) \( (2a + b)^3 \)
(ii) \( (a - 2b)^3 \)
(iii) \( (3x - 2y)^3 \)
(iv) \( (x + 5y)^3 \)
(v) \( \left(a + \frac{1}{a}\right)^3 \)
(vi) \( \left(2a - \frac{1}{2a}\right)^3 \)
Answer:
(i) \( (2a + b)^3 = (2a)^3 + (b)^3 + 3 \times 2a \times b(2a + b) \)
Using the identity \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \):
\( = 8a^3 + b^3 + 6ab(2a + b) \)
\( = 8a^3 + b^3 + 12a^2b + 6ab^2 \)

(ii) \( (a - 2b)^3 = (a)^3 - (2b)^3 - 3 \times a \times 2b(a - 2b) \)
Using the identity \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \):
\( = a^3 - 8b^3 - 6ab(a - 2b) \)
\( = a^3 - 8b^3 - 6a^2b + 12ab^2 \)

(iii) \( (3x - 2y)^3 = (3x)^3 - (2y)^3 - 3 \times 3x \times 2y(3x - 2y) \)
\( = 27x^3 - 8y^3 - 18xy(3x - 2y) \)
\( = 27x^3 - 8y^3 - 54x^2y + 36xy^2 \)

(iv) \( (x + 5y)^3 = (x)^3 + (5y)^3 + 3 \times x \times 5y(x + 5y) \)
\( = x^3 + 125y^3 + 15xy(x + 5y) \)
\( = x^3 + 125y^3 + 15x^2y + 75xy^2 \)

(v) \( \left(a + \frac{1}{a}\right)^3 = a^3 + \left(\frac{1}{a}\right)^3 + 3 \times a \times \frac{1}{a}\left(a + \frac{1}{a}\right) \)
\( = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
\( = a^3 + \frac{1}{a^3} + 3a + \frac{3}{a} \)

(vi) \( \left(2a - \frac{1}{2a}\right)^3 = (2a)^3 - \left(\frac{1}{2a}\right)^3 - 3 \times 2a \times \frac{1}{2a}\left(2a - \frac{1}{2a}\right) \)
\( = 8a^3 - \frac{1}{8a^3} - 3\left(2a - \frac{1}{2a}\right) \)
\( = 8a^3 - \frac{1}{8a^3} - 6a + \frac{3}{2a} \)
In simple words: Expand each expression using the cubic identities. Ensure that coefficients and variables are cubed properly, and expand the final product term step-by-step.

Exam Tip: Be mindful of signs when expanding \( (a-b)^3 \). Only the final term \( 3ab^2 \) ends up positive because squaring \( -b \) yields a positive value.

Question 5. (i) Find the cube of: \( a+2 \)
Answer:
By applying the algebraic identity for the cube of a binomial, \( (x+y)^3 = x^3 + y^3 + 3xy(x+y) \):
\( (a+2)^3 = (a)^3 + (2)^3 + 3 \times a \times 2(a+2) \)

\( \implies (a+2)^3 = a^3 + 8 + 6a(a+2) \)

\( \implies (a+2)^3 = a^3 + 8 + 6a^2 + 12a \)

\( \implies (a+2)^3 = a^3 + 6a^2 + 12a + 8 \)
In simple words: Expand the cube by applying the binomial formula and simplify by multiplying the terms inside the parenthesis.

Exam Tip: Keep the coefficients and exponents organized when multiplying out binomial terms to prevent calculation errors.

Question 5. (ii) Find the cube of: \( 2a-1 \)
Answer:
By using the binomial subtraction identity for cubes, \( (x-y)^3 = x^3 - y^3 - 3xy(x-y) \):
\( (2a-1)^3 = (2a)^3 - (1)^3 - 3 \times 2a \times 1(2a-1) \)

\( \implies (2a-1)^3 = 8a^3 - 1 - 6a(2a-1) \)

\( \implies (2a-1)^3 = 8a^3 - 1 - 12a^2 + 6a \)

\( \implies (2a-1)^3 = 8a^3 - 12a^2 + 6a - 1 \)
In simple words: Use the subtraction identity for cubes, and then expand the brackets carefully while keeping an eye on negative signs.

Exam Tip: Be very careful when distributing a negative sign, such as \( -6a \times -1 \), which becomes positive \( +6a \).

Question 5. (iii) Find the cube of: \( 2a+3b \)
Answer:
By utilizing the expansion identity \( (x+y)^3 = x^3 + y^3 + 3xy(x+y) \):
\( (2a+3b)^3 = (2a)^3 + (3b)^3 + 3 \times 2a \times 3b(2a+3b) \)

\( \implies (2a+3b)^3 = 8a^3 + 27b^3 + 18ab(2a+3b) \)

\( \implies (2a+3b)^3 = 8a^3 + 27b^3 + 36a^2b + 54ab^2 \)

\( \implies (2a+3b)^3 = 8a^3 + 36a^2b + 54ab^2 + 27b^3 \)
In simple words: Raise each part of the expression to the third power, calculate the middle products, and add all terms together.

Exam Tip: Remember that raising \( (2a)^3 \) to a power cubes both the number and the variable, giving you \( 8a^3 \) instead of \( 2a^3 \).

Question 5. (iv) Find the cube of: \( 3b-2a \)
Answer:
By applying the binomial cube expansion, \( (x-y)^3 = x^3 - y^3 - 3xy(x-y) \):
\( (3b-2a)^3 = (3b)^3 - (2a)^3 - 3 \times 3b \times 2a(3b-2a) \)

\( \implies (3b-2a)^3 = 27b^3 - 8a^3 - 18ab(3b-2a) \)

\( \implies (3b-2a)^3 = 27b^3 - 8a^3 - 54ab^2 + 36a^2b \)

\( \implies (3b-2a)^3 = 27b^3 - 54ab^2 + 36a^2b - 8a^3 \)
In simple words: Cube each part using the minus formula, open up the bracket, and list the final terms in descending powers of a variable.

Exam Tip: Pay attention to the order of variables in the final expansion. Keeping them in descending or ascending order is standard mathematical practice.

Question 5. (v) Find the cube of: \( 2x+\frac{1}{x} \)
Answer:
By applying the formula \( (x+y)^3 = x^3 + y^3 + 3xy(x+y) \):
\( \left(2x+\frac{1}{x}\right)^3 = (2x)^3 + \left(\frac{1}{x}\right)^3 + 3 \times 2x \times \frac{1}{x}\left(2x+\frac{1}{x}\right) \)

\( \implies \left(2x+\frac{1}{x}\right)^3 = 8x^3 + \frac{1}{x^3} + 6\left(2x+\frac{1}{x}\right) \)

\( \implies \left(2x+\frac{1}{x}\right)^3 = 8x^3 + \frac{1}{x^3} + 12x + \frac{6}{x} \)

\( \implies \left(2x+\frac{1}{x}\right)^3 = 8x^3 + 12x + \frac{6}{x} + \frac{1}{x^3} \)
In simple words: Expand using the binomial cube identity, simplify the middle term where the variables divide out, and then multiply out the rest.

Exam Tip: Be sure to simplify \( x \) and \( \frac{1}{x} \) when they multiply together, leaving you with just a constant coefficient of 6 for the middle term.

Question 5. (vi) Find the cube of: \( x-\frac{1}{2} \)
Answer:
By applying the difference identity \( (x-y)^3 = x^3 - y^3 - 3xy(x-y) \):
\( \left(x-\frac{1}{2}\right)^3 = (x)^3 - \left(\frac{1}{2}\right)^3 - 3 \times x \times \frac{1}{2}\left(x-\frac{1}{2}\right) \)

\( \implies \left(x-\frac{1}{2}\right)^3 = x^3 - \frac{1}{8} - \frac{3x}{2}\left(x-\frac{1}{2}\right) \)

\( \implies \left(x-\frac{1}{2}\right)^3 = x^3 - \frac{1}{8} - \frac{3x^2}{2} + \frac{3x}{4} \)

\( \implies \left(x-\frac{1}{2}\right)^3 = x^3 - \frac{3x^2}{2} + \frac{3x}{4} - \frac{1}{8} \)
In simple words: Apply the difference binomial cube formula, resolve the fractional coefficients carefully, and simplify the brackets.

Exam Tip: Double-check fractional calculations such as \( -\frac{3x}{2} \times -\frac{1}{2} \) to ensure the final sign is positive and the denominator is correctly multiplied.

Exercise 12(C)

Question 1. If a+b=5 and ab = 6; find \( a^2 + b^2 \)
Answer:
Given the relationships \( a + b = 5 \) and \( ab = 6 \).
We can evaluate the expression using the basic algebraic identity:
\( (a+b)^2 = a^2 + b^2 + 2ab \)
Substituting our known values into the equation:
\( (5)^2 = a^2 + b^2 + 2 \times 6 \)

\( \implies 25 = a^2 + b^2 + 12 \)

\( \implies 25 - 12 = a^2 + b^2 \)

\( \implies 13 = a^2 + b^2 \)
Therefore, the value of \( a^2 + b^2 = 13 \).
In simple words: Square the sum of the two variables, then subtract twice their product to get the sum of their squares.

Exam Tip: Working from the squared identity is the fastest way to solve this. Always remember to isolate the target term cleanly before calculating.

Question 2. If a – b = 6 and ab = 16; find \( a^2 + b^2 \)
Answer:
Given that \( a - b = 6 \) and \( ab = 16 \).
We apply the difference expansion identity:
\( (a-b)^2 = a^2 + b^2 - 2ab \)
Plugging in the given numbers:
\( (6)^2 = a^2 + b^2 - 2 \times 16 \)

\( \implies 36 = a^2 + b^2 - 32 \)

\( \implies 36 + 32 = a^2 + b^2 \)

\( \implies 68 = a^2 + b^2 \)
Thus, the sum of the squares \( a^2 + b^2 = 68 \).
In simple words: To get the sum of the squares, square the difference of the two numbers and then add twice their product.

Exam Tip: Notice that when starting with a difference \( a-b \), you must add \( 2ab \) to the square of the difference to find \( a^2 + b^2 \).

Question 3. If \( a^2 + b^2 = 29 \) and ab = 10 ; find :
(i) a + b
(ii) a – b
Answer:
(i) To determine \( a + b \), we utilize the identity:
\( (a+b)^2 = a^2 + b^2 + 2ab \)
Substituting the given parameters into this relation:
\( (a+b)^2 = 29 + 2 \times 10 \)

\( \implies (a+b)^2 = 29 + 20 \)

\( \implies (a+b)^2 = 49 \)
By taking the square root on both sides:

\( \implies a + b = \sqrt{49} \)

\( \implies a + b = 7 \)

(ii) To determine \( a - b \), we use the alternative identity:
\( (a-b)^2 = a^2 + b^2 - 2ab \)
Substituting the known values:
\( (a-b)^2 = 29 - 2 \times 10 \)

\( \implies (a-b)^2 = 29 - 20 \)

\( \implies (a-b)^2 = 9 \)
Taking the square root on both sides:

\( \implies a - b = \sqrt{9} \)

\( \implies a - b = 3 \)
In simple words: We can find the sum or difference of two variables by taking the square root of their respective expanded square expressions.

Exam Tip: For algebra questions of this level, assume positive square roots unless a negative condition is stated for the variables.

Question 4. If \( a^2 + b^2 = 10 \) and ab = 3; find :
(i) a – b
(ii) a + b
Answer:
(i) We use the binomial difference squared identity:
\( (a-b)^2 = a^2 + b^2 - 2ab \)
Plugging in our given terms:
\( (a-b)^2 = 10 - 2 \times 3 \)

\( \implies (a-b)^2 = 10 - 6 \)

\( \implies (a-b)^2 = 4 \)
Taking the square root of both sides:

\( \implies a - b = \sqrt{4} \)

\( \implies a - b = 2 \)

(ii) We now use the sum identity:
\( (a+b)^2 = a^2 + b^2 + 2ab \)
Substituting the known values:
\( (a+b)^2 = 10 + 2 \times 3 \)

\( \implies (a+b)^2 = 10 + 6 \)

\( \implies (a+b)^2 = 16 \)
Taking the square root of both sides:

\( \implies a + b = \sqrt{16} \)

\( \implies a + b = 4 \)
In simple words: The values of the sum or difference can be obtained by computing the square roots of the respective square formulas.

Exam Tip: Be sure to write down the formula clearly before substituting numbers to keep your working neat and structured.

Question 5. If \( a+\frac{1}{a}=3 \); find \( a^2 + \frac{1}{a^2} \)
Answer:
Given that \( a + \frac{1}{a} = 3 \).
Using the identity for the square of a reciprocal sum:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)
Substituting our known parameter:
\( (3)^2 = a^2 + \frac{1}{a^2} + 2 \)

\( \implies 9 = a^2 + \frac{1}{a^2} + 2 \)

\( \implies 9 - 2 = a^2 + \frac{1}{a^2} \)

\( \implies 7 = a^2 + \frac{1}{a^2} \)
So, the sum of their squares is 7.

Alternative Method:
Squaring both sides of the initial relation directly:
\( \left(a + \frac{1}{a}\right)^2 = (3)^2 \)

\( \implies a^2 + \frac{1}{a^2} + 2 = 9 \)

\( \implies a^2 + \frac{1}{a^2} = 9 - 2 \)

\( \implies a^2 + \frac{1}{a^2} = 7 \)
In simple words: Square the sum of a number and its reciprocal, and then subtract 2 to find the sum of their squares.

Exam Tip: Remember that squaring \( a + \frac{1}{a} \) yields a constant term of 2 (since the term \( 2 \cdot a \cdot \frac{1}{a} \) simplifies to 2), so you only need to subtract 2 from the square of the given value.

Question 6. If \( a-\frac{1}{a}=4 \); find \( a^2 + \frac{1}{a^2} \)
Answer:
Given that \( a - \frac{1}{a} = 4 \).
We use the squared reciprocal difference identity:
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \)
Substituting the value into this equation:
\( (4)^2 = a^2 + \frac{1}{a^2} - 2 \)

\( \implies 16 = a^2 + \frac{1}{a^2} - 2 \)

\( \implies 16 + 2 = a^2 + \frac{1}{a^2} \)

\( \implies 18 = a^2 + \frac{1}{a^2} \)
Thus, \( a^2 + \frac{1}{a^2} = 18 \).

Alternative Method:
Directly squaring both sides of the given equation:
\( \left(a - \frac{1}{a}\right)^2 = (4)^2 \)

\( \implies a^2 + \frac{1}{a^2} - 2 = 16 \)

\( \implies a^2 + \frac{1}{a^2} = 16 + 2 \)

\( \implies a^2 + \frac{1}{a^2} = 18 \)
In simple words: To find the sum of the squares of a variable and its reciprocal when their difference is known, square the difference and then add 2.

Exam Tip: Pay close attention to the sign of the constant term: squaring a difference \( a - \frac{1}{a} \) gives \( -2 \) as the cross-product, which must be added to the other side.

Question 7. If \( a^2 + \frac{1}{a^2} = 23 \); find \( a + \frac{1}{a} \)
Answer:
We know the standard reciprocal identity:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)
Plugging in the given value of \( a^2 + \frac{1}{a^2} = 23 \):
\( \left(a + \frac{1}{a}\right)^2 = 23 + 2 \)

\( \implies \left(a + \frac{1}{a}\right)^2 = 25 \)
Taking square root of both sides:

\( \implies a + \frac{1}{a} = \sqrt{25} \)

\( \implies a + \frac{1}{a} = 5 \)
In simple words: Add 2 to the sum of the squares, and then take the square root of that result to get the sum of the original reciprocal terms.

Exam Tip: Unless specified otherwise for real numbers, typically you can report the positive square root, but keeping in mind that algebraically both \( +5 \) and \( -5 \) are valid.

Question 8. If \( a^2 + \frac{1}{a^2} = 11 \); find \( a - \frac{1}{a} \)
Answer:
By utilizing the algebraic identity:
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \)
Substituting the given value \( a^2 + \frac{1}{a^2} = 11 \) into the formula:
\( \left(a - \frac{1}{a}\right)^2 = 11 - 2 \)

\( \implies \left(a - \frac{1}{a}\right)^2 = 9 \)
Taking the square root on both sides:

\( \implies a - \frac{1}{a} = \sqrt{9} \)

\( \implies a - \frac{1}{a} = 3 \)
In simple words: Subtract 2 from the sum of the squares, and then take the square root to find the difference between the terms.

Exam Tip: When evaluating a difference like \( a - \frac{1}{a} \) from the sum of squares, always remember to subtract 2 before taking the square root.

Question 9. If \( a + b + c = 10 \) and \( a^2 + b^2 + c^2 = 38 \); find \( ab + bc + ca \)
Answer:
Given: \( a + b + c = 10 \) and \( a^2 + b^2 + c^2 = 38 \).
Using the expansion identity for three terms:
\( (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \)
This can be written as:
\( (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
Substituting our values into the formula:
\( (10)^2 = 38 + 2(ab + bc + ca) \)

\( \implies 100 = 38 + 2(ab + bc + ca) \)

\( \implies 100 - 38 = 2(ab + bc + ca) \)

\( \implies 62 = 2(ab + bc + ca) \)

\( \implies ab + bc + ca = \frac{62}{2} \)

\( \implies ab + bc + ca = 31 \)

Alternative Method:
Squaring both sides of \( a + b + c = 10 \) directly:
\( (a+b+c)^2 = (10)^2 \)

\( \implies a^2 + b^2 + c^2 + 2(ab + bc + ca) = 100 \)

\( \implies 38 + 2(ab + bc + ca) = 100 \)

\( \implies 2(ab + bc + ca) = 100 - 38 \)

\( \implies 2(ab + bc + ca) = 62 \)

\( \implies ab + bc + ca = \frac{62}{2} \)

\( \implies ab + bc + ca = 31 \)
In simple words: Square the sum of all three variables, subtract the sum of their squares, and then divide the result by 2 to find the sum of their pairwise products.

Exam Tip: Be sure to factor out the common factor of 2 in the expansion of \( (a+b+c)^2 \) to get the term \( 2(ab+bc+ca) \) clearly.

Question 10. Find \( a^2 + b^2 + c^2 \); if \( a + b + c = 9 \) and \( ab + bc + ca = 24 \)
Answer:
We are given: \( a + b + c = 9 \) and \( ab + bc + ca = 24 \).
Applying the algebraic trinomial squared formula:
\( (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
Substituting the known parameters into this relation:
\( (9)^2 = a^2 + b^2 + c^2 + 2(24) \)

\( \implies 81 = a^2 + b^2 + c^2 + 48 \)

\( \implies a^2 + b^2 + c^2 = 81 - 48 \)

\( \implies a^2 + b^2 + c^2 = 33 \)
In simple words: Subtract twice the sum of the pairwise products from the square of the sum of the three terms to find the sum of their squares.

Exam Tip: Keep your calculations simple: multiply \( 2 \times 24 \) first to get 48, then subtract it directly from 81 to get the correct value of 33.

Question 11. Find \( a + b + c \); if \( a^2 + b^2 + c^2 = 83 \) and \( ab + bc + ca = 71 \)
Answer:
Given: \( a^2 + b^2 + c^2 = 83 \) and \( ab + bc + ca = 71 \).
Using the expansion identity:
\( (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
Substituting our known parameters:
\( (a+b+c)^2 = 83 + 2 \times 71 \)

\( \implies (a+b+c)^2 = 83 + 142 \)

\( \implies (a+b+c)^2 = 225 \)
Taking square root of both sides:

\( \implies a + b + c = \sqrt{225} \)

\( \implies a + b + c = 15 \)
In simple words: Add twice the pairwise product sum to the sum of the squares, and then take the square root of that total to get the sum of the three variables.

Exam Tip: Remember that 225 is a perfect square of 15, so knowing your standard squares up to 25 will save you time during exams.

Question 12. If \( a + b = 6 \) and ab = 8; find \( a^3 + b^3 \)
Answer:
Given values: \( a + b = 6 \) and \( ab = 8 \).
By utilizing the binomial cubing formula:
\( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \)
Plugging in our given terms:
\( (6)^3 = a^3 + b^3 + 3 \times 8 \times (6) \)

\( \implies 216 = a^3 + b^3 + 144 \)

\( \implies a^3 + b^3 = 216 - 144 \)

\( \implies a^3 + b^3 = 72 \)

Alternative Method:
Using the alternative form of the sum of cubes:
\( a^3 + b^3 = (a+b)^3 - 3ab(a+b) \)
Substituting our values:
\( a^3 + b^3 = (6)^3 - 3 \times 8 \times (6) \)

\( \implies a^3 + b^3 = 216 - 144 \)

\( \implies a^3 + b^3 = 72 \)
In simple words: To find the sum of two cubes, subtract three times their product multiplied by their sum from the cube of their sum.

Exam Tip: The identity \( a^3 + b^3 = (a+b)^3 - 3ab(a+b) \) is extremely direct and less prone to algebraic mistakes than other methods.

Question 13. If a – b = 3 and ab = 10; find \( a^3 - b^3 \)
Answer:
Given values: \( a - b = 3 \) and \( ab = 10 \).
Applying the algebraic formula:
\( (a-b)^3 = a^3 - b^3 - 3ab(a-b) \)
Plugging in the known terms:
\( (3)^3 = a^3 - b^3 - 3 \times 10 \times 3 \)

\( \implies 27 = a^3 - b^3 - 90 \)

\( \implies a^3 - b^3 = 27 + 90 \)

\( \implies a^3 - b^3 = 117 \)

Alternative Method:
Expressing the relation as:
\( a^3 - b^3 = (a-b)^3 + 3ab(a-b) \)
Substituting the known values:
\( a^3 - b^3 = (3)^3 + 3 \times 10 \times (3) \)

\( \implies a^3 - b^3 = 27 + 90 \)

\( \implies a^3 - b^3 = 117 \)
In simple words: To find the difference of two cubes, add three times the product multiplied by their difference to the cube of their difference.

Exam Tip: Be careful with the signs: the formula uses \( -3ab(a-b) \), which becomes \( +3ab(a-b) \) when moved to the other side to solve for \( a^3 - b^3 \).

Question 14. Find \( a^3 + \frac{1}{a^3} \) if \( a + \frac{1}{a} = 5 \)
Answer:
Given that \( a + \frac{1}{a} = 5 \).
Using the identity for cubing reciprocal sums:
\( \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
Substituting the given value into the equation:
\( (5)^3 = a^3 + \frac{1}{a^3} + 3(5) \)

\( \implies 125 = a^3 + \frac{1}{a^3} + 15 \)

\( \implies a^3 + \frac{1}{a^3} = 125 - 15 \)

\( \implies a^3 + \frac{1}{a^3} = 110 \)
In simple words: Cube the sum of a number and its reciprocal, then subtract three times that original sum to get the sum of their cubes.

Exam Tip: In reciprocal formulas, the product term \( a \cdot \frac{1}{a} \) cancels each other out, leaving a simple coefficient of 3 for the linear term.

Question 15. Find \( a^3 - \frac{1}{a^3} \) if \( a - \frac{1}{a} = 4 \)
Answer:
Given that \( a - \frac{1}{a} = 4 \).
Using the reciprocal cubing identity:
\( \left(a - \frac{1}{a}\right)^3 = a^3 - \frac{1}{a^3} - 3\left(a - \frac{1}{a}\right) \)
Substituting the value of 4 into this relation:
\( (4)^3 = a^3 - \frac{1}{a^3} - 3(4) \)

\( \implies 64 = a^3 - \frac{1}{a^3} - 12 \)

\( \implies a^3 - \frac{1}{a^3} = 64 + 12 \)

\( \implies a^3 - \frac{1}{a^3} = 76 \)
In simple words: Cube the difference of a number and its reciprocal, then add three times that original difference to find the difference of their cubes.

Exam Tip: Keep track of the negative signs: subtracting a negative term from one side means you add it when moving it to the other side.

Question 16. If \( 2x - \frac{1}{2x} = 4 \); find :
(i) \( 4x^2 + \frac{1}{4x^2} \)
(ii) \( 8x^3 - \frac{1}{8x^3} \)
Answer:
(i) We start with the given relationship:
\( 2x - \frac{1}{2x} = 4 \)
Squaring both sides of this equation:
\( \left(2x - \frac{1}{2x}\right)^2 = (4)^2 \)

\( \implies (2x)^2 + \left(\frac{1}{2x}\right)^2 - 2 \times 2x \times \frac{1}{2x} = 16 \)

\( \implies 4x^2 + \frac{1}{4x^2} - 2 = 16 \)

\( \implies 4x^2 + \frac{1}{4x^2} = 16 + 2 \)

\( \implies 4x^2 + \frac{1}{4x^2} = 18 \)

(ii) To find the difference of the cubes, we use the cubing identity:
\( \left(2x - \frac{1}{2x}\right)^3 = (2x)^3 - \left(\frac{1}{2x}\right)^3 - 3 \times 2x \times \frac{1}{2x}\left(2x - \frac{1}{2x}\right) \)
Substitute the known value of \( 2x - \frac{1}{2x} = 4 \) into the expansion:
\( (4)^3 = 8x^3 - \frac{1}{8x^3} - 3(4) \)

\( \implies 64 = 8x^3 - \frac{1}{8x^3} - 12 \)

\( \implies 8x^3 - \frac{1}{8x^3} = 64 + 12 \)

\( \implies 8x^3 - \frac{1}{8x^3} = 76 \)
In simple words: Square both sides of the expression to find the quadratic sum, and cube both sides of the expression to solve for the cubic difference.

Exam Tip: Notice that because both terms are \( 2x \), their product is \( 2x \cdot \frac{1}{2x} = 1 \), which keeps the coefficient math simple and clean.

Question 17. If \( 3x + \frac{1}{3x} = 3 \); find :
(i) \( 9x^2 + \frac{1}{9x^2} \)
(ii) \( 27x^3 + \frac{1}{27x^3} \)
Answer:
(i) Given that:
\( 3x + \frac{1}{3x} = 3 \)
Squaring both sides of this equation:
\( \left(3x + \frac{1}{3x}\right)^2 = (3)^2 \)

\( \implies (3x)^2 + \left(\frac{1}{3x}\right)^2 + 2 \times 3x \times \frac{1}{3x} = 9 \)

\( \implies 9x^2 + \frac{1}{9x^2} + 2 = 9 \)

\( \implies 9x^2 + \frac{1}{9x^2} = 9 - 2 \)

\( \implies 9x^2 + \frac{1}{9x^2} = 7 \)

(ii) For the cubic term, we cube both sides of our initial equation:
\( \left(3x + \frac{1}{3x}\right)^3 = (3)^3 \)

\( \implies (3x)^3 + \left(\frac{1}{3x}\right)^3 + 3 \times 3x \times \frac{1}{3x}\left(3x + \frac{1}{3x}\right) = 27 \)

\( \implies 27x^3 + \frac{1}{27x^3} + 3\left(3x + \frac{1}{3x}\right) = 27 \)
Substituting the initial value of \( 3x + \frac{1}{3x} = 3 \) into this relation:

\( \implies 27x^3 + \frac{1}{27x^3} + 3(3) = 27 \)

\( \implies 27x^3 + \frac{1}{27x^3} + 9 = 27 \)

\( \implies 27x^3 + \frac{1}{27x^3} = 27 - 9 \)

\( \implies 27x^3 + \frac{1}{27x^3} = 18 \)
In simple words: To find the sum of squares, square the term and subtract 2. To get the sum of cubes, cube the term and subtract three times its value.

Exam Tip: Be consistent with your algebraic formulas for sums of squares and cubes: a sum of reciprocals always simplifies the middle cross-product term to a constant.

Question 18. The sum of the squares of two numbers is 13 and their product is 6. Find:
(i) the sum of the two numbers.
(ii) the difference between them.
Answer:
Let \( x \) and \( y \) represent the two numbers. Therefore, we have:
\( x^2 + y^2 = 13 \) and \( xy = 6 \)

(i) Using the algebraic identity for the square of a sum:
\( (x + y)^2 = x^2 + y^2 + 2xy \)
Substitute the given values into the identity:
\( (x + y)^2 = 13 + 2(6) \)
\( (x + y)^2 = 13 + 12 \)
\( (x + y)^2 = 25 \)
Taking the square root on both sides:
\( x + y = \pm\sqrt{25} \)

\( \implies x + y = \pm 5 \)

(ii) Using the algebraic identity for the square of a difference:
\( (x - y)^2 = x^2 + y^2 - 2xy \)
Substitute the given values:
\( (x - y)^2 = 13 - 2(6) \)
\( (x - y)^2 = 13 - 12 \)
\( (x - y)^2 = 1 \)
Taking the square root on both sides:
\( x - y = \pm\sqrt{1} \)

\( \implies x - y = \pm 1 \)
In simple words: We use algebraic formulas to connect the sum of squares and the product of two numbers. By substituting the values, we find the squared sum is 25 and the squared difference is 1, giving us the final values after taking square roots.

Exam Tip: Remember to include both positive and negative values (\( \pm \)) when taking the square root, as squares of both positive and negative numbers are positive.

Exercise 12(D)

Question 1. Evaluate:
(i) \( (3x + \frac{1}{2})(2x + \frac{1}{3}) \)
(ii) \( (2a + 0.5)(7a - 0.3) \)
(iii) \( (9 - y)(7 + y) \)
(iv) \( (2 - z)(15 - z) \)
(v) \( (a^2 + 5)(a^2 - 3) \)
(vi) \( (4 - ab)(8 + ab) \)
(vii) \( (5xy - 7)(7xy + 9) \)
(viii) \( (3a^2 - 4b^2)(8a^2 - 3b^2) \)
Answer:
(i) Multiplying the two binomial expressions:
\( (3x + \frac{1}{2})(2x + \frac{1}{3}) = 3x(2x + \frac{1}{3}) + \frac{1}{2}(2x + \frac{1}{3}) \)
\( = 6x^2 + x + x + \frac{1}{6} \)
\( = 6x^2 + 2x + \frac{1}{6} \)

(ii) Expanding the expression term by term:
\( (2a + 0.5)(7a - 0.3) = 2a(7a - 0.3) + 0.5(7a - 0.3) \)
\( = 14a^2 - 0.6a + 3.5a - 0.15 \)
\( = 14a^2 + 2.9a - 0.15 \)

(iii) Expanding the product of binomials:
\( (9 - y)(7 + y) = 9(7 + y) - y(7 + y) \)
\( = 63 + 9y - 7y - y^2 \)
\( = 63 + 2y - y^2 \)

(iv) Distributing the terms to expand:
\( (2 - z)(15 - z) = 2(15 - z) - z(15 - z) \)
\( = 30 - 2z - 15z + z^2 \)
\( = 30 - 17z + z^2 \)

(v) Multiplying the terms with square exponents:
\( (a^2 + 5)(a^2 - 3) = a^2(a^2 - 3) + 5(a^2 - 3) \)
\( = a^4 - 3a^2 + 5a^2 - 15 \)
\( = a^4 + 2a^2 - 15 \)

(vi) Expanding the algebraic product:
\( (4 - ab)(8 + ab) = 4(8 + ab) - ab(8 + ab) \)
\( = 32 + 4ab - 8ab - a^2b^2 \)
\( = 32 - 4ab - a^2b^2 \)

(vii) Multiplying the terms step-by-step:
\( (5xy - 7)(7xy + 9) = 5xy(7xy + 9) - 7(7xy + 9) \)
\( = 35x^2y^2 + 45xy - 49xy - 63 \)
\( = 35x^2y^2 - 4xy - 63 \)

(viii) Expanding using term distribution:
\( (3a^2 - 4b^2)(8a^2 - 3b^2) = 3a^2(8a^2 - 3b^2) - 4b^2(8a^2 - 3b^2) \)
\( = 24a^4 - 9a^2b^2 - 32a^2b^2 + 12b^4 \)
\( = 24a^4 - 41a^2b^2 + 12b^4 \)
In simple words: To multiply two brackets, we multiply each term in the first bracket by every term in the second bracket, and then combine the like terms together.

Exam Tip: Pay close attention to signs when multiplying terms, especially negative signs like \( -y(7 + y) \), to avoid sign errors in the final expression.

Question 2. Evaluate:
(i) \( (2x - \frac{3}{5})(2x + \frac{3}{5}) \)
(ii) \( (\frac{4}{7}a + \frac{3}{4}b)(\frac{4}{7}a - \frac{3}{4}b) \)
(iii) \( (6 - 5xy)(6 + 5xy) \)
(iv) \( (2a + \frac{1}{2a})(2a - \frac{1}{2a}) \)
(v) \( (4x^2 - 5y^2)(4x^2 + 5y^2) \)
(vi) \( (1.6x + 0.7y)(1.6x - 0.7y) \)
(vii) \( (m + 3)(m - 3)(m^2 + 9) \)
(viii) \( (3x + 4y)(3x - 4y)(9x^2 + 16y^2) \)
(ix) \( (a + bc)(a - bc)(a^2 + b^2c^2) \)
(x) \( 203 \times 197 \)
(xi) \( 20.8 \times 19.2 \)
Answer:
These expressions can be evaluated using the difference of squares identity: \( (a - b)(a + b) = a^2 - b^2 \).

(i) Applying the difference of squares:
\( (2x - \frac{3}{5})(2x + \frac{3}{5}) = (2x)^2 - (\frac{3}{5})^2 \)
\( = 4x^2 - \frac{9}{25} \)

(ii) Using the identity:
\( (\frac{4}{7}a + \frac{3}{4}b)(\frac{4}{7}a - \frac{3}{4}b) = (\frac{4}{7}a)^2 - (\frac{3}{4}b)^2 \)
\( = \frac{16}{49}a^2 - \frac{9}{16}b^2 \)

(iii) Expanding with the identity:
\( (6 - 5xy)(6 + 5xy) = (6)^2 - (5xy)^2 \)
\( = 36 - 25x^2y^2 \)

(iv) Applying the difference of squares:
\( (2a + \frac{1}{2a})(2a - \frac{1}{2a}) = (2a)^2 - (\frac{1}{2a})^2 \)
\( = 4a^2 - \frac{1}{4a^2} \)

(v) Using the identity:
\( (4x^2 - 5y^2)(4x^2 + 5y^2) = (4x^2)^2 - (5y^2)^2 \)
\( = 16x^4 - 25y^4 \)

(vi) Applying the identity on decimal values:
\( (1.6x + 0.7y)(1.6x - 0.7y) = (1.6x)^2 - (0.7y)^2 \)
\( = 2.56x^2 - 0.49y^2 \)

(vii) First evaluate the first two terms:
\( (m + 3)(m - 3)(m^2 + 9) = [(m)^2 - (3)^2](m^2 + 9) \)
\( = (m^2 - 9)(m^2 + 9) \)
Now apply the identity again:
\( = (m^2)^2 - (9)^2 \)
\( = m^4 - 81 \)

(viii) First simplify the binomial product:
\( (3x + 4y)(3x - 4y)(9x^2 + 16y^2) = [(3x)^2 - (4y)^2](9x^2 + 16y^2) \)
\( = (9x^2 - 16y^2)(9x^2 + 16y^2) \)
Applying the identity once more:
\( = (9x^2)^2 - (16y^2)^2 \)
\( = 81x^4 - 256y^4 \)

(ix) Grouping the first two factors:
\( (a + bc)(a - bc)(a^2 + b^2c^2) = [(a)^2 - (bc)^2](a^2 + b^2c^2) \)
\( = (a^2 - b^2c^2)(a^2 + b^2c^2) \)
Now applying the identity to the resulting product:
\( = (a^2)^2 - (b^2c^2)^2 \)
\( = a^4 - b^4c^4 \)

(x) Expressing numbers relative to 200:
\( 203 \times 197 = (200 + 3)(200 - 3) \)
\( = (200)^2 - (3)^2 \)
\( = 40000 - 9 \)
\( = 39991 \)

(xi) Expressing decimals relative to 20:
\( 20.8 \times 19.2 = (20 + 0.8)(20 - 0.8) \)
\( = (20)^2 - (0.8)^2 \)
\( = 400 - 0.64 \)
\( = 399.36 \)
In simple words: When we multiply the sum and difference of the same two terms, the result is simply the square of the first term minus the square of the second term. We can also use this to multiply numbers easily by rewriting them around a round number.

Exam Tip: Recognizing the difference of squares identity \( (a-b)(a+b) = a^2-b^2 \) can save you a lot of time in algebraic expansions and arithmetic multiplications.

Question 3. Find the square of:
(i) \( 3x + \frac{2}{y} \)
(ii) \( \frac{5a}{6b} - \frac{6b}{5a} \)
(iii) \( 2m^2 - \frac{2}{3}n^2 \)
(iv) \( 5x + \frac{1}{5x} \)
(v) \( 8x + \frac{3}{2}y \)
(vi) \( 607 \)
(vii) \( 391 \)
(viii) \( 9.7 \)
Answer:
To find the squares, we use the identities \( (A + B)^2 = A^2 + 2AB + B^2 \) and \( (A - B)^2 = A^2 - 2AB + B^2 \).

(i) Applying the formula for \( (A + B)^2 \):
\( (3x + \frac{2}{y})^2 = (3x)^2 + 2(3x)(\frac{2}{y}) + (\frac{2}{y})^2 \)
\( = 9x^2 + \frac{12x}{y} + \frac{4}{y^2} \)

(ii) Applying the formula for \( (A - B)^2 \):
\( (\frac{5a}{6b} - \frac{6b}{5a})^2 = (\frac{5a}{6b})^2 - 2(\frac{5a}{6b})(\frac{6b}{5a}) + (\frac{6b}{5a})^2 \)
\( = \frac{25a^2}{36b^2} - 2 + \frac{36b^2}{25a^2} \)

(iii) Expanding using \( (A - B)^2 \):
\( (2m^2 - \frac{2}{3}n^2)^2 = (2m^2)^2 - 2(2m^2)(\frac{2}{3}n^2) + (\frac{2}{3}n^2)^2 \)
\( = 4m^4 - \frac{8}{3}m^2n^2 + \frac{4}{9}n^4 \)

(iv) Expanding using \( (A + B)^2 \):
\( (5x + \frac{1}{5x})^2 = (5x)^2 + 2(5x)(\frac{1}{5x}) + (\frac{1}{5x})^2 \)
\( = 25x^2 + 2 + \frac{1}{25x^2} \)

(v) Applying the formula for \( (A + B)^2 \):
\( (8x + \frac{3}{2}y)^2 = (8x)^2 + 2(8x)(\frac{3}{2}y) + (\frac{3}{2}y)^2 \)
\( = 64x^2 + 24xy + \frac{9}{4}y^2 \)

(vi) Rewriting 607 as \( 600 + 7 \) to expand:
\( (607)^2 = (600 + 7)^2 \)
\( = (600)^2 + 2(600)(7) + (7)^2 \)
\( = 360000 + 8400 + 49 \)
\( = 368449 \)

(vii) Rewriting 391 as \( 400 - 9 \) to expand:
\( (391)^2 = (400 - 9)^2 \)
\( = (400)^2 - 2(400)(9) + (9)^2 \)
\( = 160000 - 7200 + 81 \)
\( = 152881 \)

(viii) Rewriting 9.7 as \( 10 - 0.3 \) to expand:
\( (9.7)^2 = (10 - 0.3)^2 \)
\( = (10)^2 - 2(10)(0.3) + (0.3)^2 \)
\( = 100 - 6 + 0.09 \)
\( = 94.09 \)
In simple words: To find the square of a sum or difference of two numbers, we square the first term, add or subtract twice the product of both terms, and add the square of the second term. This same method lets us square large numbers easily by breaking them into simpler parts.

Exam Tip: When squaring a term like \( 3x \), remember to square both the coefficient and the variable, which results in \( 9x^2 \), not \( 3x^2 \).

Question 4. If \( a + \frac{1}{a} = 2 \), find:
(i) \( a^2 + \frac{1}{a^2} \)
(ii) \( a^4 + \frac{1}{a^4} \)
Answer:
Given that \( a + \frac{1}{a} = 2 \).

(i) Squaring both sides of the given equation:
\( (a + \frac{1}{a})^2 = 2^2 \)
\( a^2 + 2(a)(\frac{1}{a}) + \frac{1}{a^2} = 4 \)
\( a^2 + 2 + \frac{1}{a^2} = 4 \)
Subtracting 2 from both sides:
\( a^2 + \frac{1}{a^2} = 4 - 2 \)

\( \implies a^2 + \frac{1}{a^2} = 2 \)

(ii) To find the fourth power, we square the result obtained in part (i):
\( (a^2 + \frac{1}{a^2})^2 = 2^2 \)
\( a^4 + 2(a^2)(\frac{1}{a^2}) + \frac{1}{a^4} = 4 \)
\( a^4 + 2 + \frac{1}{a^4} = 4 \)
Subtracting 2 from both sides:
\( a^4 + \frac{1}{a^4} = 4 - 2 \)

\( \implies a^4 + \frac{1}{a^4} = 2 \)
In simple words: To find higher even powers when given \( a + \frac{1}{a} \), we square the expression we already know and then subtract 2 from the result.

Exam Tip: Keep in mind that \( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \). The product of the term and its reciprocal always simplifies to a constant, which is \( 2 \).

Question 5. If \( m - \frac{1}{m} = 5 \), find:
(i) \( m^2 + \frac{1}{m^2} \)
(ii) \( m^4 + \frac{1}{m^4} \)
(iii) \( m^2 - \frac{1}{m^2} \)
Answer:
Given that \( m - \frac{1}{m} = 5 \).

(i) Squaring both sides of the given equation:
\( (m - \frac{1}{m})^2 = 5^2 \)
\( m^2 - 2(m)(\frac{1}{m}) + \frac{1}{m^2} = 25 \)
\( m^2 - 2 + \frac{1}{m^2} = 25 \)
Adding 2 to both sides:
\( m^2 + \frac{1}{m^2} = 25 + 2 \)

\( \implies m^2 + \frac{1}{m^2} = 27 \)

(ii) Squaring the result from part (i) to get the fourth powers:
\( (m^2 + \frac{1}{m^2})^2 = 27^2 \)
\( m^4 + 2(m^2)(\frac{1}{m^2}) + \frac{1}{m^4} = 729 \)
\( m^4 + 2 + \frac{1}{m^4} = 729 \)
Subtracting 2 from both sides:
\( m^4 + \frac{1}{m^4} = 729 - 2 \)

\( \implies m^4 + \frac{1}{m^4} = 727 \)

(iii) Using the identity for difference of squares:
\( m^2 - \frac{1}{m^2} = (m + \frac{1}{m})(m - \frac{1}{m}) \)
We already know that \( m - \frac{1}{m} = 5 \). Now we need to find \( m + \frac{1}{m} \).
Using the algebraic relationship:
\( (m + \frac{1}{m})^2 = (m - \frac{1}{m})^2 + 4 \)
\( (m + \frac{1}{m})^2 = 5^2 + 4 \)
\( (m + \frac{1}{m})^2 = 25 + 4 = 29 \)
Taking the positive square root:
\( m + \frac{1}{m} = \sqrt{29} \)
Substituting these values back into the difference of squares product:
\( m^2 - \frac{1}{m^2} = 5 \times \sqrt{29} \)

\( \implies m^2 - \frac{1}{m^2} = 5\sqrt{29} \)
In simple words: When given a minus term, squaring it and adding 2 gives the sum of squares. Squaring that result again and subtracting 2 gives the sum of fourth powers. To find the difference of squares, we multiply the sum and difference of the simple terms together.

Exam Tip: Remember the crucial identity that links the square of a sum and a difference: \( (A + B)^2 = (A - B)^2 + 4AB \). It is very useful for converting sum terms to difference terms quickly.

Question 6. If \( a^2 + b^2 = 41 \) and \( ab = 4 \), find:
(i) \( a - b \)
(ii) \( a + b \)
Answer:
Given that \( a^2 + b^2 = 41 \) and \( ab = 4 \).

(i) Using the identity for the square of a difference:
\( (a - b)^2 = a^2 + b^2 - 2ab \)
Substitute the given values:
\( (a - b)^2 = 41 - 2(4) \)
\( (a - b)^2 = 41 - 8 \)
\( (a - b)^2 = 33 \)
Taking the positive square root:
\( a - b = \sqrt{33} \)

(ii) Using the identity for the square of a sum:
\( (a + b)^2 = a^2 + b^2 + 2ab \)
Substitute the given values:
\( (a + b)^2 = 41 + 2(4) \)
\( (a + b)^2 = 41 + 8 \)
\( (a + b)^2 = 49 \)
Taking the square root on both sides:
\( a + b = \sqrt{49} \)

\( \implies a + b = 7 \)
In simple words: To find the sum or difference of two numbers when we know their squares and product, we expand the formulas for \( (a+b)^2 \) and \( (a-b)^2 \), plug in our values, and then take the square roots of the results.

Exam Tip: Unless specified otherwise, taking a square root can mathematically result in a positive or negative value, but for many standard textbook algebra problems, the positive root is often the principal focus.

Question 7. If \( 2a + \frac{1}{2a} = 8 \), find:
(i) \( 4a^2 + \frac{1}{4a^2} \)
(ii) \( 16a^4 + \frac{1}{16a^4} \)
Answer:
Given that \( 2a + \frac{1}{2a} = 8 \).

(i) Squaring both sides of the given equation:
\( (2a + \frac{1}{2a})^2 = 8^2 \)
\( 4a^2 + 2(2a)(\frac{1}{2a}) + \frac{1}{4a^2} = 64 \)
\( 4a^2 + 2 + \frac{1}{4a^2} = 64 \)
Subtracting 2 from both sides:
\( 4a^2 + \frac{1}{4a^2} = 64 - 2 \)

\( \implies 4a^2 + \frac{1}{4a^2} = 62 \)

(ii) Squaring the result obtained in part (i) to find the fourth powers:
\( (4a^2 + \frac{1}{4a^2})^2 = 62^2 \)
\( 16a^4 + 2(4a^2)(\frac{1}{4a^2}) + \frac{1}{16a^4} = 3844 \)
\( 16a^4 + 2 + \frac{1}{16a^4} = 3844 \)
Subtracting 2 from both sides:
\( 16a^4 + \frac{1}{16a^4} = 3844 - 2 \)

\( \implies 16a^4 + \frac{1}{16a^4} = 3842 \)
In simple words: Squaring the binomial cancels out the reciprocal terms, leaving a constant 2. We subtract this constant to isolate our target squared expression, repeating this step to get to higher powers.

Exam Tip: Always calculate the square of two-digit numbers like \( 62^2 \) carefully. Working it out as \( (60+2)^2 = 3600 + 240 + 4 = 3844 \) is a quick mental check to ensure accuracy.

Question 8. If \( 3x - \frac{1}{3x} = 5 \), find:
(i) \( 9x^2 + \frac{1}{9x^2} \)
(ii) \( 81x^4 + \frac{1}{81x^4} \)
Answer:
Given that \( 3x - \frac{1}{3x} = 5 \).

(i) Squaring both sides of the given equation:
\( (3x - \frac{1}{3x})^2 = 5^2 \)
\( 9x^2 - 2(3x)(\frac{1}{3x}) + \frac{1}{9x^2} = 25 \)
\( 9x^2 - 2 + \frac{1}{9x^2} = 25 \)
Adding 2 to both sides:
\( 9x^2 + \frac{1}{9x^2} = 25 + 2 \)

\( \implies 9x^2 + \frac{1}{9x^2} = 27 \)

(ii) Squaring the result obtained in part (i) to find the fourth powers:
\( (9x^2 + \frac{1}{9x^2})^2 = 27^2 \)
\( 81x^4 + 2(9x^2)(\frac{1}{9x^2}) + \frac{1}{81x^4} = 729 \)
\( 81x^4 + 2 + \frac{1}{81x^4} = 729 \)
Subtracting 2 from both sides:
\( 81x^4 + \frac{1}{81x^4} = 729 - 2 \)

\( \implies 81x^4 + \frac{1}{81x^4} = 727 \)
In simple words: Since we have a minus sign initially, squaring the term leaves us with a minus 2. We add 2 to move it to the other side. To get to the fourth power, we square again, which leaves a plus 2, so we subtract 2.

Exam Tip: Be careful with the signs: squaring a negative binomial like \( (A - \frac{1}{A})^2 \) introduces a \( -2 \) constant, whereas squaring a positive binomial like \( (A^2 + \frac{1}{A^2})^2 \) introduces a \( +2 \) constant.

Question 9. Expand:
(i) \( (3x - 4y + 5z)^2 \)
(ii) \( (2a - 5b - 4c)^2 \)
(iii) \( (5x + 3y)^3 \)
(iv) \( (6a - 7b)^3 \)
Answer:
We expand these expressions using algebraic identities.

(i) Using the identity \( (A + B + C)^2 = A^2 + B^2 + C^2 + 2AB + 2BC + 2CA \):
\( (3x - 4y + 5z)^2 = (3x)^2 + (-4y)^2 + (5z)^2 + 2(3x)(-4y) + 2(-4y)(5z) + 2(5z)(3x) \)
\( = 9x^2 + 16y^2 + 25z^2 - 24xy - 40yz + 30zx \)

(ii) Applying the same trinomial identity:
\( (2a - 5b - 4c)^2 = (2a)^2 + (-5b)^2 + (-4c)^2 + 2(2a)(-5b) + 2(-5b)(-4c) + 2(-4c)(2a) \)
\( = 4a^2 + 25b^2 + 16c^2 - 20ab + 40bc - 16ca \)

(iii) Using the binomial cube identity \( (A + B)^3 = A^3 + B^3 + 3AB(A + B) \):
\( (5x + 3y)^3 = (5x)^3 + (3y)^3 + 3(5x)(3y)(5x + 3y) \)
\( = 125x^3 + 27y^3 + 45xy(5x + 3y) \)
\( = 125x^3 + 27y^3 + 225x^2y + 135xy^2 \)

(iv) Using the binomial cube difference identity \( (A - B)^3 = A^3 - B^3 - 3AB(A - B) \):
\( (6a - 7b)^3 = (6a)^3 - (7b)^3 - 3(6a)(7b)(6a - 7b) \)
\( = 216a^3 - 343b^3 - 126ab(6a - 7b) \)
\( = 216a^3 - 343b^3 - 756a^2b + 882ab^2 \)
Rearranging the terms in descending powers of \( a \):
\( = 216a^3 - 756a^2b + 882ab^2 - 343b^3 \)
In simple words: We expand these multi-term algebraic expressions using direct formulas for squaring three terms or cubing two terms. We make sure to distribute negative signs properly during multiplication.

Exam Tip: Be very careful when squaring a negative term like \( (-5b)^2 \), which becomes a positive \( 25b^2 \), whereas the cross-product terms like \( 2(2a)(-5b) \) will carry a negative sign.

Question 10. If \( a + b + c = 9 \) and \( ab + bc + ca = 15 \), find: \( a^2 + b^2 + c^2 \)
Answer:
We use the trinomial expansion identity:
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
Substituting the values given in the question:
\( (9)^2 = a^2 + b^2 + c^2 + 2(15) \)
\( 81 = a^2 + b^2 + c^2 + 30 \)
Isolating the sum of squares:
\( a^2 + b^2 + c^2 = 81 - 30 \)

\( \implies a^2 + b^2 + c^2 = 51 \)
In simple words: The square of the sum of three numbers is equal to the sum of their squares plus twice the sum of their pairwise products. Using this formula, we easily find the unknown sum of squares.

Exam Tip: Memorize the three-variable identity \( (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \) as it frequently appears in algebraic evaluation questions.

Question 11. If \( a + b + c = 11 \) and \( a^2 + b^2 + c^2 = 81 \), find \( ab + bc + ca \).
Answer:
Applying the trinomial expansion identity:
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
Substituting the known values:
\( (11)^2 = 81 + 2(ab + bc + ca) \)
\( 121 = 81 + 2(ab + bc + ca) \)
Subtracting 81 from both sides:
\( 2(ab + bc + ca) = 121 - 81 \)
\( 2(ab + bc + ca) = 40 \)
Dividing by 2:
\( ab + bc + ca = \frac{40}{2} \)

\( \implies ab + bc + ca = 20 \)
In simple words: We plug the known values into the trinomial expansion formula. By subtracting the sum of squares from the squared total, and then dividing the remainder by 2, we find the sum of their products.

Exam Tip: Be careful with simple arithmetic operations like subtraction and division here to avoid dropping easy marks on final calculation steps.

Question 12. If \( 3x - 4y = 5 \) and \( xy = 3 \), find: \( 27x^3 - 64y^3 \).
Answer:
We can write the target expression as the difference of two cubes:
\( 27x^3 - 64y^3 = (3x)^3 - (4y)^3 \)
Using the identity \( A^3 - B^3 = (A - B)^3 + 3AB(A - B) \), where \( A = 3x \) and \( B = 4y \):
\( (3x)^3 - (4y)^3 = (3x - 4y)^3 + 3(3x)(4y)(3x - 4y) \)
\( = (3x - 4y)^3 + 36xy(3x - 4y) \)
Substitute the given values \( 3x - 4y = 5 \) and \( xy = 3 \) into the expression:
\( = (5)^3 + 36(3)(5) \)
\( = 125 + 540 \)

\( \implies 27x^3 - 64y^3 = 665 \)
In simple words: We rewrite the expression as a difference of cubes. We then use an identity that lets us express this difference using only the given values, which we plug in to get the final answer.

Exam Tip: Ensure that you correctly multiply the coefficients when computing \( 3AB \), which gives \( 3 \times 3x \times 4y = 36xy \), before substituting the value of \( xy \).

Question 13. If \( a + b = 8 \) and \( ab = 15 \), find: \( a^3 + b^3 \).
Answer:
Using the algebraic identity for the sum of two cubes:
\( a^3 + b^3 = (a + b)^3 - 3ab(a + b) \)
Substituting the given values into the identity:
\( a^3 + b^3 = (8)^3 - 3(15)(8) \)
\( = 512 - 360 \)

\( \implies a^3 + b^3 = 152 \)
In simple words: The sum of cubes can be calculated directly by cubing the sum of the numbers and subtracting three times their product multiplied by their sum.

Exam Tip: Memorizing basic cubes like \( 8^3 = 512 \) is highly recommended for quick and accurate calculations under exam pressure.

Question 14. If \( 3x + 2y = 9 \) and \( xy = 3 \), find: \( 27x^3 + 8y^3 \).
Answer:
We express the target terms as a sum of cubes:
\( 27x^3 + 8y^3 = (3x)^3 + (2y)^3 \)
Using the identity \( A^3 + B^3 = (A + B)^3 - 3AB(A + B) \) with \( A = 3x \) and \( B = 2y \):
\( (3x)^3 + (2y)^3 = (3x + 2y)^3 - 3(3x)(2y)(3x + 2y) \)
\( = (3x + 2y)^3 - 18xy(3x + 2y) \)
Substituting the given values \( 3x + 2y = 9 \) and \( xy = 3 \) into the equation:
\( = (9)^3 - 18(3)(9) \)
\( = 729 - 486 \)

\( \implies 27x^3 + 8y^3 = 243 \)
In simple words: To find the sum of cubes of these terms, we cube their total sum and subtract the product term calculated from their individual parts.

Exam Tip: Be sure to write the product term as \( 18xy \) (derived from \( 3 \times 3x \times 2y \)), rather than simply using \( 3xy \) which is a common formula misapplication.

Question 15. If \( 5x - 4y = 7 \) and \( xy = 8 \), find: \( 125x^3 - 64y^3 \).
Answer:
We can write the target expression as:
\( 125x^3 - 64y^3 = (5x)^3 - (4y)^3 \)
Using the identity \( A^3 - B^3 = (A - B)^3 + 3AB(A - B) \) where \( A = 5x \) and \( B = 4y \):
\( (5x)^3 - (4y)^3 = (5x - 4y)^3 + 3(5x)(4y)(5x - 4y) \)
\( = (5x - 4y)^3 + 60xy(5x - 4y) \)
Substituting the values given in the problem:
\( = (7)^3 + 60(8)(7) \)
\( = 343 + 3360 \)

\( \implies 125x^3 - 64y^3 = 3703 \)
In simple words: This problem is solved by recognizing the terms as cubes and using the difference-of-cubes formula to relate them back to our given linear values.

Exam Tip: Be careful with large multiplication steps like \( 60 \times 8 \times 7 = 3360 \) to ensure that you do not make simple computational errors.

Question 16. The difference between two numbers is 5 and their products is 14. Find the difference between their cubes.
Answer:
Let \( x \) and \( y \) be the two numbers. From the given conditions:
\( x - y = 5 \) and \( xy = 14 \)
We need to find the difference between their cubes, which is \( x^3 - y^3 \).
Using the identity:
\( x^3 - y^3 = (x - y)^3 + 3xy(x - y) \)
Substitute the known values into this equation:
\( x^3 - y^3 = (5)^3 + 3(14)(5) \)
\( = 125 + 210 \)

\( \implies x^3 - y^3 = 335 \)
In simple words: We translate the word problem into two equations. Then, we use the standard algebraic identity for the difference of two cubes to directly compute the required value.

Exam Tip: Carefully read word problems to set up the algebraic equations correctly before deciding which algebraic identities to apply.