RS Aggarwal Class 8 Mathematics Solutions Chapter 4 Cubes and Cube Roots

 

Exercise 4A

 

Question 1. Find the cube of 8, 15, 21, and 60.
Answer: (i) \( (8)^3 = (8 \times 8 \times 8) = 512 \). Thus, the cube of 8 is 512.
(ii) \( (15)^3 = (15 \times 15 \times 15) = 3375 \). Thus, the cube of 15 is 3375.
(iii) \( (21)^3 = (21 \times 21 \times 21) = 9261 \). Thus, the cube of 21 is 9261.
(iv) \( (60)^3 = (60 \times 60 \times 60) = 216000 \). Thus, the cube of 60 is 216000.
In simple words: Multiply a number by itself three times to find its cube. For instance, 8 cubed means 8 times 8 times 8, which gives 512.

Exam Tip: Always multiply step by step to avoid calculation errors when finding cubes of larger numbers.

 

Question 2. Find the cube of 1.2, 3.5, 0.8, and 0.05.
Answer: (i) \( (1.2)^3 = (1.2 \times 1.2 \times 1.2) = 1.728 \). Thus, the cube of 1.2 is 1.728.
(ii) \( (3.5)^3 = (3.5 \times 3.5 \times 3.5) = 42.875 \). Thus, the cube of 3.5 is 42.875.
(iii) \( (0.8)^3 = (0.8 \times 0.8 \times 0.8) = 0.512 \). Thus, the cube of 0.8 is 0.512.
(iv) \( (0.05)^3 = (0.05 \times 0.05 \times 0.05) = 0.000125 \). Thus, the cube of 0.05 is 0.000125.
In simple words: To cube a decimal number, multiply it by itself three times just like you would with whole numbers. Pay careful attention to where the decimal point goes in the final answer.

Exam Tip: Count decimal places carefully - if a number has 1 decimal place, its cube will have 3 decimal places.

 

Question 3. Find the cube of \( \frac{2}{3} \), \( \frac{10}{11} \), \( \frac{1}{16} \), and \( 1\frac{3}{10} \).
Answer: (i) \( \left(\frac{2}{3}\right)^3 = \left(\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}\right) = \left(\frac{64}{343}\right) \). Thus, the cube of \( \frac{2}{3} \) is \( \frac{64}{343} \).

(ii) \( \left(\frac{10}{11}\right)^3 = \left(\frac{10}{11} \times \frac{10}{11} \times \frac{10}{11}\right) = \left(\frac{1000}{1331}\right) \). Thus, the cube of \( \frac{10}{11} \) is \( \frac{1000}{1331} \).

(iii) \( \left(\frac{1}{16}\right)^3 = \left(\frac{1}{16} \times \frac{1}{16} \times \frac{1}{16}\right) = \left(\frac{1}{3376}\right) \). Thus, the cube of \( \frac{1}{16} \) is \( \frac{1}{3376} \). Also, \( 1\frac{3}{10} = \frac{13}{10} \). \( \left(\frac{13}{10}\right)^3 = \left(\frac{13}{10} \times \frac{13}{10} \times \frac{13}{10}\right) = \left(\frac{2197}{1000}\right) \). Thus, the cube of \( 1\frac{3}{10} \) is \( \frac{2197}{1000} \).
In simple words: For fractions, cube both the numerator and denominator separately. A mixed number should first be converted to an improper fraction, then cubed.

Exam Tip: Always convert mixed numbers to improper fractions before cubing to avoid mistakes.

 

Question 4. Which of the following are perfect cubes? 125, 243, 343, 256, 8000, 9261, 5324, 3375.
Answer: (i) 125 - Resolving 125 into prime factors: \( 125 = 5 \times 5 \times 5 \). Here, one triplet is formed, which is \( 5^3 \). Hence, 125 can be written as the product of the triplets of 5. Therefore, 125 is a perfect cube.

(ii) 243 - 243 is not a perfect cube.

(iii) 343 - Resolving 343 into prime factors: \( 343 = 7 \times 7 \times 7 \). Here, one triplet is formed, which is \( 7^3 \). Hence, 343 can be written as the product of the triplets of 7. Therefore, 343 is a perfect cube.

(iv) 256 - 256 is not a perfect cube.

(v) 8000 - Resolving 8000 into prime factors: \( 8000 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \). Here, three triplets are formed, which are \( 2^3 \), \( 2^3 \), and \( 5^3 \). Hence, 8000 can be written as the product of the triplets of 2, 2, and 5, i.e. \( 2^3 \times 2^3 \times 5^3 = 20^3 \). Therefore, 8000 is a perfect cube.

(vi) 9261 - Resolving 9261 into prime factors: \( 9261 = 3 \times 3 \times 3 \times 7 \times 7 \times 7 \). Here, two triplets are formed, which are \( 3^3 \) and \( 7^3 \). Hence, 9261 can be written as the product of the triplets of 3 and 7, i.e. \( 3^3 \times 7^3 = 21^3 \). Therefore, 9261 is a perfect cube.

(vii) 5324 - 5324 is not a perfect cube.

(viii) 3375 - Resolving 3375 into prime factors: \( 3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 \). Here, two triplets are formed, which are \( 3^3 \) and \( 5^3 \). Hence, 3375 can be written as the product of the triplets of 3 and 5, i.e. \( 3^3 \times 5^3 = 15^3 \). Therefore, 3375 is a perfect cube.
In simple words: A perfect cube is a number that can be expressed as a triplet (group of three identical factors) after breaking it into prime factors. If all prime factors form complete triplets with nothing left over, it is a perfect cube.

Exam Tip: Always factorize completely and group the prime factors into triplets - if any prime appears a number of times that is not a multiple of 3, the number is not a perfect cube.

 

Question 5. Show that the cube of an even number is even and the cube of an odd number is odd.
Answer: The cubes of even numbers are always even. Therefore, 216, 512, and 1000 are the cubes of even numbers. \( 216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3 = 6^3 \). \( 512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^3 \times 2^3 \times 2^3 = 8^3 \). \( 1000 = 2 \times 2 \times 5 \times 5 \times 5 = 2^3 \times 5^3 = 10^3 \). The cube of an odd number is an odd number. Therefore, 125, 343, and 9261 are the cubes of odd numbers. \( 125 = 5 \times 5 \times 5 = 5^3 \). \( 343 = 7 \times 7 \times 7 = 7^3 \). \( 9261 = 3 \times 3 \times 3 \times 7 \times 7 \times 7 = 3^3 \times 7^3 = 21^3 \).
In simple words: When you cube an even number (a number divisible by 2), the result is always even because all prime factors include at least one 2, making the product even. When you cube an odd number (not divisible by 2), the result stays odd because no factor of 2 is introduced.

Exam Tip: Remember that even × even = even, and odd × odd = odd. Since cubing means multiplying a number three times, these rules apply throughout the multiplication chain.

 

Question 6. By what least number should 1323 be multiplied to make it a perfect cube?
Answer: For 1323: Breaking 1323 into prime factors gives us: \( 3 | 1323 \), \( 3 | 441 \), \( 3 | 147 \), \( 7 | 49 \), \( 7 | 7 \), and 1 at the end. So \( 1323 = 3 \times 3 \times 3 \times 7 \times 7 \). To make it a perfect cube, it has to be multiplied by 7. Thus, \( 1323 \times 7 = 9261 = 3^3 \times 7^3 = 21^3 \).
In simple words: Decompose the number into its prime factors. Then identify which prime factors don't form a complete triplet. Multiply by the missing factors to complete the triplets.

Exam Tip: Count the frequency of each prime factor - if a factor appears once or twice (not three times), you need to multiply by it to reach three occurrences.

 

Question 7. By what least number should 2560 be divided to make it a perfect square?
Answer: For 2560: Breaking 2560 into prime factors gives us: \( 2 | 2560 \), \( 2 | 1280 \), \( 2 | 640 \), \( 2 | 320 \), \( 2 | 160 \), \( 2 | 80 \), \( 2 | 40 \), \( 2 | 20 \), \( 2 | 10 \), \( 5 | 5 \), and 1 at the end. So \( 2560 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \). To make this a perfect square, we have to multiply it by \( 5 \times 5 \).
In simple words: Look at the prime factorization. For a perfect square, every prime must appear an even number of times. Here, the number 5 appears only once, so we need one more 5. We multiply by 5 × 5 to complete the pairing.

Exam Tip: Note that the question asks about making a perfect square, not a cube - check the exponent requirement (2 for squares, 3 for cubes).

 

Question 8. By what least number should 1600 be divided to make it a perfect cube?
Answer: For 1600: Breaking 1600 into prime factors gives us: \( 2 | 1600 \), \( 2 | 800 \), \( 2 | 400 \), \( 2 | 200 \), \( 2 | 100 \), \( 2 | 50 \), \( 5 | 25 \), and \( 5 | 5 \). So \( 1600 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \). Therefore, to make the quotient a perfect cube, we have to divide 1600 by \( 5 \times 5 = 25 \).
In simple words: Find the prime factors of the number. For a perfect cube, each prime factor must appear a number of times that is a multiple of 3. Divide by the excess factors to reach the nearest multiple of 3 for each prime.

Exam Tip: Identify which prime factors are "over" the nearest multiple of 3 - those are the ones you must divide out.

 

Question 9. By what least number should 8788 be divided so that the quotient is a perfect cube?
Answer: For 8788: Breaking 8788 into prime factors gives us: \( 2 | 8788 \), \( 2 | 4394 \), \( 13 | 2197 \), \( 13 | 169 \), and \( 13 | 13 \). So \( 8788 = 2 \times 2 \times 13 \times 13 \times 13 \). Therefore, 8788 should be divided by 4, i.e. \( (2 \times 2) \), so that the quotient is a perfect cube.
In simple words: Break the number down into its prime factors. Any prime that does not appear three times (or a multiple of three times) needs to be removed. Divide by those extra factors.

Exam Tip: Group the prime factors into triplets - any leftover factors (those that don't form a complete triplet) are what you divide out.

 

Exercise 4B

 

Question 1. Find \( (2 + 5)^3 \) using the formula \( a^3 + 3a^2b + 3ab^2 + b^3 \).
Answer: Here, \( a = 2 \) and \( b = 5 \). Using the formula \( a^3 + 3a^2b + 3ab^2 + b^3 \):

Therefore, \( (2 + 5)^3 = 15625 \).
In simple words: The formula breaks down the cube of a sum into four manageable parts: the cube of the first term, three times the square of the first times the second, three times the first times the square of the second, and the cube of the second term. Add them up to get your answer.

Exam Tip: Use the algebraic formula when numbers are difficult to cube directly - it often simplifies the arithmetic involved.

 

Question 2. Find \( (4 + 7)^3 \) using the formula \( a^3 + 3a^2b + 3ab^2 + b^3 \).
Answer: Here, \( a = 4 \) and \( b = 7 \). Using the formula \( a^3 + 3a^2b + 3ab^2 + b^3 \):

Therefore, \( (4 + 7)^3 = 103823 \).
In simple words: Apply each component of the formula step by step. The formula lets you avoid computing (4 + 7)³ directly as 11³, which is trickier to calculate mentally.

Exam Tip: Double-check your arithmetic in each cell of the table before adding the final results together.

 

Question 3. Find \( (6 + 8)^3 \) using the formula \( a^3 + 3a^2b + 3ab^2 + b^3 \).
Answer: Here, \( a = 6 \) and \( b = 8 \). Using the formula \( a^3 + 3a^2b + 3ab^2 + b^3 \):

Therefore, \( (6 + 8)^3 = 314432 \).
In simple words: Follow the same systematic approach for each term in the formula. Breaking a compound operation into simpler steps makes the calculation more manageable and less error-prone.

Exam Tip: Organize your work in a table format to keep track of all components and avoid losing terms during addition.

 

Question 4. Find \( (8 + 4)^3 \) using the formula \( a^3 + 3a^2b + 3ab^2 + b^3 \).
Answer: Here, \( a = 8 \) and \( b = 4 \). Using the formula \( a^3 + 3a^2b + 3ab^2 + b^3 \):

Therefore, \( (8 + 4)^3 = 592704 \).
In simple words: Plug your values for \( a \) and \( b \) into each part of the formula. The systematic breakdown prevents mistakes that might occur if you tried to calculate (8 + 4)³ = 12³ as a single step.

Exam Tip: Verify your answer by computing 12³ separately to check if your formula result is correct.

 

Exercise 4C

 

Question 1. Find \( \sqrt[3]{64} \).
Answer: By prime factorization: \( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \). Therefore, \( \sqrt[3]{64} = \sqrt[3]{(2)^3 \times (2)^3} = (2 \times 2) = 4 \).
In simple words: Break the number into its prime factors. Group them in sets of three (triplets). Take one number from each triplet and multiply them together to get the cube root.

Exam Tip: Always organize prime factors into triplets - this makes it clear which factors to extract for the cube root.

 

Question 2. Find \( \sqrt[3]{343} \).
Answer: By prime factorization: \( 343 = 7 \times 7 \times 7 = (7 \times 7 \times 7) \). Therefore, \( \sqrt[3]{343} = \sqrt[3]{7^3} = 7 \).
In simple words: When a prime appears exactly three times in the factorization, the cube root is simply that prime. The factorization is already in the form of one perfect triplet.

Exam Tip: Recognize when a number is a perfect cube - it has all its primes in groups of exactly three.

 

Question 3. Find \( \sqrt[3]{729} \).
Answer: By prime factorization: \( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) \). Therefore, \( \sqrt[3]{729} = (3 \times 3) = 9 \).
In simple words: Divide the prime factors into triplets. The cube root is the product of one factor taken from each triplet.

Exam Tip: Count the total number of prime factors - for a perfect cube, it should be divisible by 3.

 

Question 4. Find \( \sqrt[3]{1728} \).
Answer: By prime factorization: \( 1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \). Therefore, \( \sqrt[3]{1728} = (2 \times 2 \times 3) = 12 \).
In simple words: Once you have organized the prime factors into triplets, multiply one representative from each triplet to find the cube root.

Exam Tip: Write out the grouping of triplets explicitly - this visual method reduces calculation mistakes.

 

Question 5. Find \( \sqrt[3]{9261} \).
Answer: By prime factorization: \( 9261 = 3 \times 3 \times 3 \times 7 \times 7 \times 7 = (3 \times 3 \times 3) \times (7 \times 7 \times 7) \). Therefore, \( \sqrt[3]{9261} = (3 \times 7) = 21 \).
In simple words: Factorize the number completely. Create groups of three identical factors. From each group, pick one factor and multiply all of them to find the cube root.

Exam Tip: If you can't form triplets of all prime factors, then the number is not a perfect cube and doesn't have a whole number cube root.

 

Question 6. Find \( \sqrt[3]{4096} \).
Answer: By prime factorization: \( 4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 2^3 \times 2^3 \times 2^3 \times 2^3 \). Therefore, \( \sqrt[3]{4096} = (2 \times 2 \times 2 \times 2) = 16 \).
In simple words: Count how many times each prime factor appears. Divide that count by 3 to find how many of that prime goes into the cube root.

Exam Tip: A quick method: if a prime \( p \) appears \( n \) times, it appears \( n/3 \) times in the cube root.

 

Question 7. Find \( \sqrt[3]{8000} \).
Answer: By prime factorization: \( 8000 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5) \). Therefore, \( \sqrt[3]{8000} = (2 \times 2 \times 5) = 20 \).
In simple words: The cube root extracts one factor from each triplet. If you see 2 appearing 6 times, it contributes 2 × 2 = 4 to the cube root (since 6 ÷ 3 = 2).

Exam Tip: To verify, cube your answer: 20³ should equal 8000.

 

Question 8. Find \( \sqrt[3]{3375} \).
Answer: By prime factorization: \( 3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 = (3 \times 3 \times 3) \times (5 \times 5 \times 5) \). Therefore, \( \sqrt[3]{3375} = (3 \times 5) = 15 \).
In simple words: Every triplet of identical primes contributes one of those primes to the cube root. Multiply all such contributions together.

Exam Tip: For numbers with two different prime factors, the cube root will be the product of those primes (each appearing once).

 

Question 9. Find \( \sqrt[3]{-216} \).
Answer: By prime factorization: \( 216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \). Therefore, \( \sqrt[3]{-216} = -(2 \times 3) = -6 \). Also, \( \sqrt[3]{-216} = -(\sqrt[3]{216}) = -6 \).
In simple words: The cube root of a negative number is also negative. First find the cube root of the positive version, then apply the negative sign.

Exam Tip: Unlike square roots, cube roots of negative numbers are defined and yield negative results.

 

Question 10. Find \( \sqrt[3]{-512} \).
Answer: By prime factorization: \( 512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \). Therefore, \( \sqrt[3]{-512} = -(2 \times 2 \times 2) = -8 \). Also, \( \sqrt[3]{-512} = -(\sqrt[3]{512}) = -8 \).
In simple words: The cube root of a negative number follows the same process as for a positive number, except the final result carries a negative sign.

Exam Tip: Remember that \( (-a)^3 = -a^3 \), so \( \sqrt[3]{-a} = -\sqrt[3]{a} \).

 

Question 11. Find \( \sqrt[3]{-1331} \).
Answer: By prime factorization: \( \sqrt[3]{1331} = \sqrt[3]{11 \times 11 \times 11} \). From the prime factorization, \( 11 | 1331 \), \( 11 | 121 \), \( 11 | 11 \), and 1 at the end. Therefore, \( \sqrt[3]{-1331} = -(11 \times 11 \times 11)^{1/3} = -11 \). Also, \( \sqrt[3]{-1331} = -(\sqrt[3]{1331}) = -11 \).
In simple words: Factor the number into primes. Since 11 appears three times, the cube root is simply 11. For the negative number, add the negative sign to get -11.

Exam Tip: When you see a repeated prime that appears three or six or nine times, you can quickly identify the cube root.

 

Question 12. Find \( \sqrt[3]{\frac{27}{64}} \).
Answer: By prime factorization: For 27: \( 3 | 27 \), \( 3 | 9 \), \( 3 | 3 \), and 1 at the end. For 64: \( 2 | 64 \), \( 2 | 32 \), \( 2 | 16 \), \( 2 | 8 \), \( 2 | 4 \), \( 2 | 2 \), and 1 at the end. Therefore, \( \sqrt[3]{\frac{27}{64}} = \frac{\sqrt[3]{27}}{\sqrt[3]{64}} = \frac{\sqrt[3]{(3 \times 3 \times 3)}}{\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2)}} = \frac{\sqrt[3]{(3)^3}}{\sqrt[3]{(4)^4}} = \frac{3}{4} \). Therefore, \( \sqrt[3]{\frac{27}{64}} = \frac{3}{4} \).
In simple words: The cube root of a fraction equals the cube root of the numerator divided by the cube root of the denominator. Factor each separately and simplify.

Exam Tip: Always factorize the numerator and denominator independently before taking the cube root of the final fraction.

 

Question 13. Find \( \sqrt[3]{\frac{125}{216}} \).
Answer: By prime factorization: For 125: \( 5 | 125 \), \( 5 | 25 \), \( 5 | 5 \), and 1 at the end. For 216: \( 2 | 216 \), \( 2 | 108 \), \( 2 | 54 \), \( 3 | 27 \), \( 3 | 9 \), \( 3 | 3 \), and 1 at the end. Therefore, \( \sqrt[3]{\frac{125}{216}} = \frac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)}} = \frac{\sqrt[3]{(5)^3}}{\sqrt[3]{(6)^3}} = \frac{5}{6} \). Therefore, \( \sqrt[3]{\frac{125}{216}} = \frac{5}{6} \).
In simple words: Break both numerator and denominator into prime factors. Once you identify the triplets, take one from each triplet and form the cube root numerator and denominator.

Exam Tip: Check if the numerator and denominator are both perfect cubes - if so, the cube root will be a simple fraction with whole numbers.

 

Question 14. Find \( \sqrt[3]{\frac{27}{125}} \).
Answer: By factorization: \( 27 = 3 \times 3 \times 3 \) and \( 125 = 5 \times 5 \times 5 \). Therefore, \( \sqrt[3]{\frac{27}{125}} = \frac{\sqrt[3]{3 \times 3 \times 3}}{\sqrt[3]{5 \times 5 \times 5}} = \frac{3}{5} \).
In simple words: Both 27 and 125 are perfect cubes, so their cube roots are simply 3 and 5 respectively, giving you the fraction 3/5.

Exam Tip: Memorize the cubes of small numbers (2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125, etc.) to speed up fraction cube root problems.

 

Question 15. Find \( \sqrt[3]{\frac{64}{343}} \).
Answer: On factorization: \( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \) and \( 343 = 7 \times 7 \times 7 \). Therefore, \( \sqrt[3]{\frac{64}{343}} = \frac{\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2}}{\sqrt[3]{7 \times 7 \times 7}} = \frac{4}{7} \).
In simple words: 64 has six factors of 2, so the cube root has two factors of 2. 343 is the cube of 7. Combine them to get 4/7.

Exam Tip: When the numerator is a perfect cube with a different prime than the denominator, the calculation becomes straightforward.

 

Question 16. Find \( \sqrt[3]{64 \times 729} \).
Answer: \( \sqrt[3]{64 \times 729} = \sqrt[3]{64} \times \sqrt[3]{729} = \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2)} \times \sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3)} = (4) \times (9) = 36 \).
In simple words: You can separate the cube root of a product into the product of the individual cube roots. This often makes the calculation easier.

Exam Tip: Remember that \( \sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} \) - this property helps break down complex problems.

 

Question 17. Find \( \sqrt[3]{\frac{729}{1000}} \).
Answer: On factorization: For 729: \( 3 | 729 \), \( 3 | 243 \), \( 3 | 81 \), \( 3 | 27 \), \( 3 | 9 \), \( 3 | 3 \), and 1 at the end. For 1000: \( 2 | 1000 \), \( 2 | 500 \), \( 2 | 250 \), \( 5 | 125 \), \( 5 | 25 \), \( 5 | 5 \), and 1 at the end. On factorization: \( \sqrt[3]{\frac{729}{1000}} = \frac{\sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3)}}{\sqrt[3]{(2 \times 2 \times 2) \times (5 \times 5 \times 5)}} = \frac{(3 \times 3)}{(2 \times 5)} = \frac{9}{10} \). Therefore, \( \sqrt[3]{\frac{729}{1000}} = \frac{9}{10} \).
In simple words: Identify how many triplets each prime has in both numerator and denominator. One number from each triplet goes into the cube root.

Exam Tip: Write the factorization in grouped form to make it obvious how many triplets you have.

 

Question 18. Find \( \sqrt[3]{\frac{512}{343}} \).
Answer: By factorization: \( 512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \) and \( 343 = 7 \times 7 \times 7 \). Therefore, \( \sqrt[3]{\frac{512}{343}} = \frac{\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}}{\sqrt[3]{7 \times 7 \times 7}} = \frac{(2 \times 2 \times 2) \times (2 \times 2 \times 2)}{7} = \frac{8}{7} \).
In simple words: 512 has nine factors of 2, which form three triplets, so the cube root has three factors of 2. 343 is 7³, so the cube root is 7. Combine them to get 8/7.

Exam Tip: Always verify: (8/7)³ should equal 512/343 when you cube it back.

 

Exercise 4D

 

Question 1. Which of the following are perfect cubes? (a) 141 (b) 294 (c) 216 (d) 496.
Answer: (a) 141 is not a perfect cube.

(b) 294 is not a perfect cube.

(c) (\(\checkmark\)) 216 is a perfect cube. \( 216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) = \left(2^3\right) \times \left(3^3\right) = 6^3 \).

(d) 496 is not a perfect cube.
In simple words: A number is a perfect cube only if, after breaking it into prime factors, every prime appears a number of times that is divisible by 3. In this case, only 216 qualifies.

Exam Tip: Quickly eliminate numbers that have prime factors appearing 1, 2, 4, 5, etc. times (any count not divisible by 3).

 

Question 2. Which of the following are perfect cubes? (a) 1152 (b) 1331 (c) 2016 (d) 739.
Answer: (a) \( 1152 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = \left(2\right)^3 \times \left(2\right)^3 \times (2 \times 3 \times 3) \). Hence, 1152 is not a perfect cube.

(b) (\(\checkmark\)) \( 1331 = 11 \times 11 \times 11 = \left(11\right)^3 \). Hence, 1331 is a perfect cube.

(c) \( 2016 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 = \left(2\right)^3 \times 2 \times 3 \times 3 \times 7 \). Hence, 2016 is not a perfect cube.

(d) 739 is not a perfect cube.
In simple words: Only 1331 breaks down into three elevens multiplied together, making it the only perfect cube in this set.

Exam Tip: If one prime factor doesn't divide evenly into groups of 3, mark it as "not a perfect cube" immediately.

 

Question 3. By what least number should 1944 be multiplied to make it a perfect cube?
Answer: The answer is to be generated based on the prime factorization of 1944 and identification of which prime factors need to be completed to form triplets for a perfect cube.
In simple words: Factorize 1944, look at which primes don't form complete triplets, and multiply by the minimum needed to complete them all.

Exam Tip: Write out the prime factorization step-by-step so you don't miss any factors.

 

Question 4. Find \( \sqrt[3]{512} \).
Answer: (c) 8. \( \sqrt[3]{512} = \sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2} = \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)} = \sqrt[3]{\left(2\right)^3 \times \left(2\right)^3 \times \left(2\right)^3} = 8 \). Hence, the cube root of 512 is 8.
In simple words: 512 has nine factors of 2, which form three complete triplets. Taking one 2 from each triplet gives 2 × 2 × 2 = 8.

Exam Tip: Once you recognize the triplet pattern, extracting the cube root becomes straightforward.

 

Question 5. Find \( \sqrt[3]{\frac{64}{343}} \).
Answer: (b) \( \frac{4}{7} \). \( \sqrt[3]{\frac{64}{343}} = \frac{\sqrt[3]{64}}{\sqrt[3]{343}} = \frac{\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2)}}{\sqrt[3]{7 \times 7 \times 7}} = \frac{4}{7} \). Therefore, \( \sqrt[3]{\frac{64}{343}} = \frac{4}{7} \).
In simple words: 64 breaks into two triplets of 2, giving 2 × 2 = 4 in the cube root. 343 is 7³, giving 7 in the cube root. The final answer is 4/7.

Exam Tip: Test your answer by cubing it back: (4/7)³ should equal 64/343.

 

Question 6. Find \( \sqrt[3]{\frac{-512}{729}} \).
Answer: (b) \( -\frac{8}{9} \). \( \sqrt[3]{\frac{-512}{729}} = \frac{\sqrt[3]{(-8) \times (-8) \times (-8)}}{\sqrt[3]{9 \times 9 \times 9}} = \frac{(-8)}{(9)} \). Therefore, \( \sqrt[3]{\frac{-512}{729}} = -\frac{8}{9} \).
In simple words: The cube root of a negative fraction is negative. Find the cube root of the positive version (512/729 = 8/9), then apply the negative sign.

Exam Tip: Negative numbers under a cube root always yield negative results - this is different from square roots.

 

Question 7. By what least number should 648 be multiplied to make it a perfect cube?
Answer: (c) 9. \( 648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 = \left(2\right)^3 \times \left(3\right)^3 \times 3 \). Therefore, to get a perfect cube, we need to multiply 648 by 9, i.e. \( (3 \times 3) \).
In simple words: In the prime factorization of 648, the number 3 appears four times. To make it a triplet (three times), we need one more 3. Actually, we need two more 3s to get six total (which is 3 × 3 = 9).

Exam Tip: Count the excess factors beyond the nearest multiple of 3 for each prime - those excesses determine what you must multiply by.

 

Question 8. By what least number should 1536 be divided to make it a perfect cube?
Answer: (a) 3. \( 1536 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = \left(2\right)^3 \times \left(2\right)^3 \times \left(2\right)^3 \times 3 \). Therefore, to get a perfect cube, we need to divide 1536 by 3.
In simple words: 1536 has nine factors of 2 (three complete triplets) and only one factor of 3. Dividing out that single 3 leaves only the triplets, making the result a perfect cube.

Exam Tip: Identify primes that appear fewer than three times - divide them out completely to achieve a perfect cube.

 

Question 9. By what least number should 1536 be divided to make it a perfect cube?
Answer: (a) 3. \( 1536 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = \left(2\right)^3 \times \left(2\right)^3 \times \left(2\right)^3 \times 3 \). Therefore, to get a perfect cube, we need to divide 1536 by 3.
In simple words: Since 1536 contains nine 2s (forming three complete triplets) and only one 3, removing the 3 by division gives you a perfect cube of 2.

Exam Tip: Always complete the prime factorization first - this shows exactly what to divide out.

 

Question 10. Find the value of \( (0.8)^3 \).
Answer: (c) 0.512. \( (0.8)^3 = (0.8) \times (0.8) \times (0.8) = 0.512 \). Therefore, \( (0.8)^3 = 0.512 \).
In simple words: Multiply 0.8 by itself three times. The result has three decimal places (since 0.8 has one, and 1 × 3 = 3).

Exam Tip: For decimals, count decimal places: if the base has \( n \) decimal places, the cube has \( 3n \) decimal places.