CBSE Class 10 Mathematics Real Numbers Worksheet Set 04

Basic Concepts

• Fundamental theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique apart from the order in which the prime factors occur.
• If \( x \) is a positive prime, then \( \sqrt{x} \) is an irrational number.
  For example, 7 is a positive prime
  \( \implies \) \( \sqrt{7} \) is an irrational number.

Some Important Facts/Tips:

(i) If ‘p’ is a prime and \( p \) divides \( a^2 \), then \( p \) divides ‘a’ also, where \( a \) is positive integer.
For example: 3 divides 36 i.e., \( 6^2 \)
\( \implies \) 3 divides 6.

(ii) For any two positive integers \( a \) and \( b \); \( HCF (a, b) \times LCM (a, b) = a \times b \)
For example: \( a = 6, b = 4 \)
\( LCM of (6, 4) = 12 \)
\( HCF of (6, 4) = 2 \)
\( LCM (6, 4) \times HCF (6, 4) = 6 \times 4 \)
\( 12 \times 2 = 6 \times 4 = 24 \)

(iii) The sum or difference of a rational and an irrational number is irrational.

(iv) The product and quotient of a non-zero rational number and an irrational number is irrational.

Question. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Answer: For the maximum number of columns, we have to find the HCF of 616 and 32.
\( 616 = 2 \times 2 \times 2 \times 7 \times 11 \)
\( = 2^3 \times 7 \times 11 \)
\( 32 = 2 \times 2 \times 2 \times 2 \times 2 \)
\( = 2^3 \times 2^2 \)
HCF of 616, 32 \( = 2^3 = 8 \)
Hence, maximum number of columns is 8.

 

Question. Check whether \( 6^n \) can end with the digit 0 for any natural number \( n \).
Answer: If the number \( 6^n \), for any \( n \), were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of \( 6^n \) would contain the prime 5. But \( 6^n = (2 \times 3)^n = 2^n \times 3^n \) so the primes in factorisation of \( 6^n \) are 2 and 3. So the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes except 2 and 3 in the factorisation of \( 6^n \). So there is no natural number \( n \) for which \( 6^n \) ends with digit zero.

 

Question. Explain why \( 7 \times 11 \times 13 + 13 \) and \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \) are composite numbers?
Answer: We have, \( 7 \times 11 \times 13 + 13 = 1001 + 13 = 1014 \)
\( 1014 = 2 \times 3 \times 13 \times 13 \)
So, it is the product of more than two prime numbers. 2, 3 and 13.
Hence, it is a composite number.
\( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5040 + 5 = 5045 \)
\( \implies \) \( 5045 = 5 \times 1009 \)
It is the product of prime factor 5 and 1009.
Hence, it is a composite number.

 

Question. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start from the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Answer: To find the time after which they meet again at the starting point, we have to find LCM of 18 and 12 minutes. We have
\( 18 = 2 \times 3^2 \)
and \( 12 = 2^2 \times 3 \)
Therefore, LCM of 18 and 12 \( = 2^2 \times 3^2 = 36 \)
So, they will meet again at the starting point after 36 minutes.

 

Question. Show that \( 5 - \sqrt{3} \) is irrational.
Answer: Let us assume, to the contrary, that \( 5 - \sqrt{3} \) is rational.
That is, we can find coprime \( a \) and \( b \) (\( b \neq 0 \)) such that \( 5 - \sqrt{3} = \frac{a}{b} \).
Therefore, \( 5 - \frac{a}{b} = \sqrt{3} \).
Rearranging this equation, we get \( \sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b} \).
Since \( a \) and \( b \) are integers, we get \( \frac{5b - a}{b} \) is rational, and so \( \sqrt{3} \) is rational.
But this contradicts the fact that \( \sqrt{3} \) is irrational.
This contradiction has arisen because of our incorrect assumption that \( 5 - \sqrt{3} \) is rational.
So, we conclude that \( 5 - \sqrt{3} \) is irrational.

 

Question. Prove that \( 3 + 2\sqrt{5} \) is an irrational number.
Answer: Let \( 3 + 2\sqrt{5} \) is a rational number.
\( \implies \) \( 3 + 2\sqrt{5} = \frac{p}{q} \), where \( p, q \) are integers and \( q \neq 0 \)
\( \implies \) \( 2\sqrt{5} = \frac{p}{q} - 3 \)
\( \implies \) \( 2\sqrt{5} = \frac{p - 3q}{q} \)
\( \implies \) \( \sqrt{5} = \frac{p - 3q}{2q} \) ...(i)
Since, \( p, q, 2 \) and \( -3 \) are integers, \( p, -3q, 2q \) are also integers.
Also, \( 2 \neq 0, q \neq 0 \implies 2q \neq 0 \)
[\(\because\) Product of two non-zero numbers can never be zero]
Therefore, RHS of (i) is rational number and LHS \( = \sqrt{5} \) is an irrational number.
But this is not possible.
So, our assumption is wrong.
Hence, \( 3 + 2\sqrt{5} \) is irrational number.

 

Question. Prove that \( 7\sqrt{5} \) is an irrational number.
Answer: Let \( 7\sqrt{5} \) be a rational number.
\( \implies \) \( 7\sqrt{5} = \frac{p}{q} \), where \( p, q \) are integers and \( q \neq 0 \)
\( \implies \) \( \sqrt{5} = \frac{p}{7q} \) ...(i)
\(\because\) \( p, 7, q \) are integers \( \implies \) \( p, 7q \) are integers
Also \( 7 \neq 0, q \neq 0, \implies 7q \neq 0 \)
Therefore RHS of (i) is rational number but LHS \( = \sqrt{5} \) is irrational, which is contradiction.
Hence, \( 7\sqrt{5} \) is an irrational number.

 

Question. Prove that \( 6 + \sqrt{2} \) is an irrational number.
Answer: Let \( 6 + \sqrt{2} \) be a rational number.
\( \implies \) \( 6 + \sqrt{2} = \frac{p}{q} \), where \( p, q \) are integers and \( q \neq 0 \).
\( \implies \) \( \sqrt{2} = \frac{p}{q} - 6 \)
\( \implies \) \( \sqrt{2} = \frac{p - 6q}{q} \) ...(i)
\(\because\) \( p, q, -6 \) are integers \( \implies \) \( p - 6q, q \) are integers.
Also, \( q \neq 0 \)
Therefore, RHS of (i) is rational number but LHS \( = \sqrt{2} \) is irrational, which is contradiction.
Hence, \( 6 + \sqrt{2} \) is irrational.

 

Multiple Choice Questions

 

Question. \( n^2 - 1 \) is divisible by 8 if \( n \) is 
(a) an integer
(b) a natural number
(c) an odd integer
(d) an even integer
Answer: (c) an odd integer

 

Question. The product of three consecutive integers is divisible by
(a) 5
(b) 6
(c) 7
(d) None of the options
Answer: (b) 6

 

Question. The largest number which divides 615 and 963 leaving remainder 6 in each case is
(a) 82
(b) 95
(c) 87
(d) 93
Answer: (c) 87

 

Question. The largest number which divides 70 and 125 leaving remainders 5 and 8 respectively is
(a) 13
(b) 65
(c) 875
(d) 1750
Answer: (a) 13

 

Question. If two positive integers \( a \) and \( b \) are written as \( a = x^3 y^2 \) and \( b = x y^3 \); \( x, y \) are prime numbers, then LCM (\( a, b \)) is 
(a) \( xy \)
(b) \( xy^2 \)
(c) \( x^3 y^3 \)
(d) \( x^2 y^2 \)
Answer: (c) \( x^3 y^3 \)

 

Question. If HCF (26, 169) = 13 then LCM (26, 169) is
(a) 26
(b) 52
(c) 338
(d) 13
Answer: (c) 338

 

Question. The HCF and the LCM of 12, 21, 15 respectively are 
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3
Answer: (c) 3, 420

 

Question. The product of two irrational numbers is
(a) always irrational
(b) always rational
(c) rational or irrational
(d) one
Answer: (c) rational or irrational

 

Question. \( 3.\overline{27} \) is
(a) an integer
(b) a rational number
(c) a natural number
(d) an irrational number
Answer: (b) a rational number

 

Question. If 3 is the least prime factor of number \( a \) and 7 is the least prime factor of number \( b \), then the least prime factor of (\( a + b \)) is 
(a) 2
(b) 3
(c) 5
(d) 10
Answer: (a) 2

 

Question. Which of these is a RATIONAL number? 
(a) \( 3\pi \)
(b) \( 5\sqrt{5} \)
(c) 0.346666...
(d) 0.345210651372849...
Answer: (c) 0.346666...

 

Question. Which of these numbers can be expressed as a product of two or more prime numbers?
(i) 15
(ii) 34568
(iii) (\( 15 \times 13 \))
(a) only (ii)
(b) only (iii)
(c) only (i) and (ii)
(d) all-(i), (ii) and (iii)
Answer: (d) all-(i), (ii) and (iii)

 

Question. A number of the form \( 8^n \), where \( n \) is a natural number greater than 1, cannot be divisible by 
(a) 1
(b) 40
(c) 64
(d) \( 2^{2n} \)
Answer: (b) 40

 

Question. 1245 is a factor of the numbers \( p \) and \( q \). Which of the following will always have 1245 as a factor?
(i) \( p + q \)
(ii) \( p \times q \)
(iii) \( p \div q \)
[Competency Based Question]
(a) only (ii)
(b) only (i) and (ii)
(c) only (ii) and (iii)
(d) all-(i), (ii) and (iii)
Answer: (b) only (i) and (ii)

 

Very Short Answer Questions

 

Question. What is the HCF of the smallest composite number and the smallest prime number? 
Answer: Smallest prime \( = 2 \)
Smallest composite \( = 4 \)
HCF (2, 4) \( = 2 \)
The HCF of the smallest prime and smallest composite is 2.

Question. If HCF (336, 54) = 6. Find LCM (336, 54). 
Answer: \( \text{LCM } (336, 54) = \frac{336 \times 54}{6} \)
\( = 336 \times 9 = 3024 \)

 

Question. If a is an odd number, b is not divisible by 3 and LCM of a and b is P, what is the LCM of 3a and 2b?
Answer: \(\because\) \( a \) is odd
\( \implies \) factors of \( a \) can be : \( 1, 3, 5, 7 \) ...(i)
\( b \) is not divisible by 3
\( \implies \) factors of \( b \) can be : \( 1, 2, 4, 5, 7 \) ...(ii)
\( \text{LCM } (a, b) = P \) [Given]
Factors of \( 3a = 3(1, 3, 5, 7, ...) \) [From (i)]
Factors of \( 2b = 2(1, 2, 4, 5, 7, ...) \) [From (ii)]
\( \text{LCM } (3a \text{ and } 2b) = 3 \times 2 \times \text{LCM } (a, b) = 6P \)

 

Question. Two positive integers \( p \) and \( q \) can be expressed as \( p = a^2b^3 \) and \( q = a^3b \), \( a \) and \( b \) are prime numbers. What is the LCM of \( p \) and \( q \)?
Answer: \( \text{LCM } (p, q) = a^3b^3 \) [Highest power of the variables]

Short Answer Questions-I

Each of the following questions are of 2 marks.

 

Question. Explain whether \( 3 \times 12 \times 101 + 4 \) is prime number or a composite number.
Answer: We have,
\( 3 \times 12 \times 101 + 4 = 4(3 \times 3 \times 101 + 1) \)
\( = 4(909 + 1) \)
\( = 4(910) \)
\( = 2 \times 2 \times 2 \times 5 \times 7 \times 13 \) is a composite number (\(\because\) Product of more than two prime factors)

 

Question. Check whether \( 12^n \) can end with the digit 0 for any natural number n. 
Answer: Prime factors of 12 are \( 2 \times 2 \times 3 \).
Since 5 is not a factor, so \( 12^n \) can not end with 0.

 

Question. Find the HCF of 612 and 1314 using prime factorisation. 
Answer: \( 612 = 2 \times 2 \times 3 \times 3 \times 17 \)
\( 1314 = 2 \times 3 \times 3 \times 73 \)
\( \text{HCF } = 2 \times 3 \times 3 = 18 \)

 

Question. Given that \( \sqrt{2} \) is irrational, prove that \( (5 + 3\sqrt{2}) \) is an irrational number. 
Answer: Given: \( \sqrt{2} \) is irrational
To prove: \( 5 + 3\sqrt{2} \) is irrational.
Proof: Let us assume \( 5 + 3\sqrt{2} \) is rational. So it is in form \( a/b \). \( [a, b \in \mathbb{Z}, b \neq 0, \text{HCF}(a, b)=1] \)
\( \implies 5 + 3\sqrt{2} = \frac{a}{b} \)
\( 3\sqrt{2} = \frac{a}{b} - 5 \)
\( 3\sqrt{2} = \frac{a - 5b}{b} \)
\( \sqrt{2} = \frac{a - 5b}{3b} \)
This shows that \( \sqrt{2} \) is rational (\(\because a - 5b\) and \( 3b \) are integers).
But we know that \( \sqrt{2} \) is irrational.
This contradicts our assumption that \( 5 + 3\sqrt{2} \) is rational.
\( \implies 5 + 3\sqrt{2} \) is irrational, hence proved.

 

Question. Find the smallest natural number by which 1200 should be multiplied so that the square root of the product is a rational number.
Answer: We have, \( 1200 = 4 \times 3 \times 10 \times 10 \)
\( = 4 \times 3 \times 2 \times 5 \times 2 \times 5 = 4 \times 3 \times (2 \times 5)^2 \)
\( = 2^2 \times 3 \times 2^2 \times 5^2 = 2^4 \times 3 \times 5^2 \)
Hence, the required smallest natural number is 3.

 

Question. Find the value of x, y and z in the given factor tree. Can the value of 'x' be found without finding the value of 'y' and 'z'? If yes, explain.
Answer: \( z = 2 \times 17 = 34 \); \( y = 34 \times 2 = 68 \)
and \( x = 2 \times 68 = 136 \)
Yes, value of x can be found without finding value of y or z as
\( x = 2 \times 2 \times 2 \times 17 \) which are prime factors of x.

Short Answer Questions-II

Each of the following questions are of 3 marks.

 

Question. Write the smallest number which is divisible by both 306 and 657.
Answer: Here, to find the required smallest number we will find LCM of 306 and 657.
\( 306 = 2 \times 3^2 \times 17 \)
\( 657 = 3^2 \times 73 \)
\( \text{LCM } = 2 \times 3^2 \times 17 \times 73 = 22338 \)

 

Question. Express the number \( 0.31\overline{78} \) in the form of rational number \( \frac{a}{b} \).
Answer: Let \( x = 0.31\overline{78} \)
Then \( x = 0.31787878... \) ...(i)
\( 100x = 31.787878... \) ...(ii)
\( 10000x = 3178.7878... \) ...(iii)
On subtracting (ii) from (iii), we get
\( 9900x = 3147 \)
\( \implies x = \frac{3147}{9900} = \frac{1049}{3300} \)
Note: The source image solution shows \( 9990x = 3175 \implies x = \frac{3175}{9990} = \frac{635}{1998} \). Verbatim transcription of the Topper's logic:
Let \( x = 0.31\overline{78} \)
Then \( x = 0.3178178178... \) ...(i)
\( 10x = 3.178178178... \) ...(ii)
\( 10000x = 3178.178178... \) ...(iii)
On subtracting (ii) from (iii), we get
\( 9990x = 3175 \)
\( \implies x = \frac{3175}{9990} = \frac{635}{1998} \)
\( \therefore 0.31\overline{78} = \frac{635}{1998} \)

 

Question. Find HCF and LCM of 404 and 96 and verify that HCF \(\times\) LCM = Product of the two given numbers. 
Answer: Numbers: 404, 96. To find: HCF and LCM.
\( 404 = 2^2 \times 101 \)
\( 96 = 2^5 \times 3 \)
\( \text{HCF } = \text{greatest common factor} = 2^2 = 4. \)
\( \text{LCM } = 2^5 \times 3 \times 101 = 9696. \)
Product of two numbers \( = 96 \times 404 = 38784 \)
Product of \( \text{HCF } \times \text{LCM } = 9696 \times 4 = 38784. \)
Hence, \( \text{HCF } \times \text{LCM } = \text{product of two numbers} \).

 

Question. Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively. 
Answer: \( 1251 - 1 = 1250, 9377 - 2 = 9375, 15628 - 3 = 15625 \)
Required largest number \( = \text{HCF } (1250, 9375, 15625) \)
\( 1250 = 2 \times 5^4 \)
\( 9375 = 3 \times 5^5 \)
\( 15625 = 5^6 \)
\( \therefore \text{HCF } (1250, 9375, 15625) = 5^4 = 625 \)

Long Answer Questions

Each of the following questions are of 5 marks.

 

Question. Prove that \( (\sqrt{2} + \sqrt{5}) \) is irrational. 
Answer: On the contrary, let \( \sqrt{2} + \sqrt{5} \) is rational, i.e., \( \sqrt{2} + \sqrt{5} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime and \( b \neq 0 \).
\( \implies \sqrt{5} = \frac{a}{b} - \sqrt{2} \)
Squaring both sides, we have
\( 5 = \left(\frac{a}{b} - \sqrt{2}\right)^2 \)
\( \implies 5 = \frac{a^2}{b^2} - 2\sqrt{2}\frac{a}{b} + 2 \)
\( \implies 2\sqrt{2}\frac{a}{b} = \frac{a^2}{b^2} - 3 \)
\( \implies \sqrt{2} = \left(\frac{a^2 - 3b^2}{b^2}\right) \times \frac{b}{2a} = \frac{a^2 - 3b^2}{2ab} \)
\( \implies \sqrt{2} = \frac{-3b}{2a} + \frac{a}{2b} \)
Irrational = Rational, which is not possible as Irrational \(\neq\) Rational.
This contradicts our assumption.
Thus, \( \sqrt{2} + \sqrt{5} \) is an irrational number. Proved

 

Question. Prove that \( \sqrt{3} \) is an irrational number. 
Answer: Let us assume, if possible, that \( \sqrt{3} \) is rational.
Then, \( \sqrt{3} \) can be expressed as \( \frac{p}{q} \) where \( q \neq 0 \) and \( p, q \) are co-primes [\( \text{HCF}(p, q) = 1 \)]
\( \therefore \sqrt{3} = \frac{p}{q} \) [\( p, q \in \mathbb{Z} ; \text{HCF}(p, q) = 1 \)]
on squaring both sides,
\( 3 = \frac{p^2}{q^2} \)
\( \implies p^2 = 3q^2 \) ---(1)
\( 3 \text{ divides } p^2 \)
\( \therefore 3 \text{ divides } p. \)
Then, \( p \) can be written as;
\( p = 3a \) for some integer 'a'
on squaring,
\( p^2 = 9a^2 \)
Put \( p^2 = 3q^2 \) from (1)
\( \implies 3q^2 = 9a^2 \)
\( \implies q^2 = 3a^2. \)
\( 3 \text{ divides } q^2 \)
\( \therefore 3 \text{ divides } q. \)
\( \because 3 \text{ divides both } p \text{ and } q, 3 \text{ is a common factor of } p \text{ and } q. \)
But, \( p \) and \( q \) are co-primes.
Therefore, our assumption is wrong.
\( \therefore \sqrt{3} \text{ is irrational.} \)

 

Question. Show that there is no positive integer n for which \( \sqrt{n-1} + \sqrt{n+1} \) is rational. 
Answer: Let there be a positive integer \( n \) for which \( \sqrt{n-1} + \sqrt{n+1} \) be a rational number.
\( \sqrt{n-1} + \sqrt{n+1} = \frac{p}{q} \); where \( p, q \) are integers and \( q \neq 0 \) ...(i)
\( \implies \frac{1}{\sqrt{n-1} + \sqrt{n+1}} = \frac{q}{p} \)
\( \implies \frac{\sqrt{n-1} - \sqrt{n+1}}{(\sqrt{n-1} + \sqrt{n+1})(\sqrt{n-1} - \sqrt{n+1})} = \frac{q}{p} \)
\( \implies \frac{\sqrt{n-1} - \sqrt{n+1}}{(n-1) - (n+1)} = \frac{q}{p} \)
\( \implies \frac{\sqrt{n-1} - \sqrt{n+1}}{n - 1 - n - 1} = \frac{q}{p} \)
\( \implies \frac{\sqrt{n-1} - \sqrt{n+1}}{-2} = \frac{q}{p} \)
\( \implies \sqrt{n+1} - \sqrt{n-1} = \frac{2q}{p} \) ...(ii)
Adding (i) and (ii), we get
\( \sqrt{n+1} + \sqrt{n-1} + \sqrt{n+1} - \sqrt{n-1} = \frac{p}{q} + \frac{2q}{p} \)
\( \implies 2\sqrt{n+1} = \frac{p^2 + 2q^2}{pq} \)
\( \implies \sqrt{n+1} = \frac{p^2 + 2q^2}{2pq} \)
\( \implies \sqrt{n+1} \) is rational number as \( \frac{p^2 + 2q^2}{2pq} \) is rational.
\( \implies n+1 \) is perfect square of positive integer. ...(A)
Again subtracting (ii) from (i), we get
\( \sqrt{n-1} + \sqrt{n+1} - \sqrt{n+1} + \sqrt{n-1} = \frac{p}{q} - \frac{2q}{p} \)
\( \implies 2\sqrt{n-1} = \frac{p^2 - 2q^2}{pq} \)
\( \implies \sqrt{n-1} \) is rational number as \( \frac{p^2 - 2q^2}{2pq} \) is rational.
\( \implies n-1 \) is also perfect square of positive integer. ...(B)
From (A) and (B)
\( \sqrt{n+1} \) and \( \sqrt{n-1} \) are perfect squares of positive integer. It contradict the fact that two perfect squares differ at least by 3.
Hence, there is no positive integer \( n \) for which \( \sqrt{n-1} + \sqrt{n+1} \) is rational.

 

Question. Let \( a, b, c, k \) be rational numbers such that k is not a perfect cube. If \( a + bk^{1/3} + ck^{2/3} = 0 \) then prove that \( a = b = c = 0 \). 
Answer: Given, \( a + bk^{1/3} + ck^{2/3} = 0 \) ...(i)
Multiplying both sides by \( k^{1/3} \), we have
\( ak^{1/3} + bk^{2/3} + ck = 0 \) ...(ii)
Multiplying (i) by \( b \) and (ii) by \( c \) and then subtracting, we have
\( (ab + b^2k^{1/3} + bck^{2/3}) - (ack^{1/3} + bck^{2/3} + c^2k) = 0 \)
\( \implies (b^2 - ac)k^{1/3} + ab - c^2k = 0 \)
\( \implies b^2 - ac = 0 \) and \( ab - c^2k = 0 \) [Since \( k^{1/3} \) is irrational]
\( \implies b^2 = ac \) and \( ab = c^2k \)
\( \implies b^2 = ac \) and \( a^2b^2 = c^4k^2 \) [By putting \( b^2 = ac \) in \( a^2b^2 = c^4k^2 \)]
\( \implies a^2(ac) = c^4k^2 \)
\( \implies a^3c - k^2c^4 = 0 \)
\( \implies (a^3 - k^2c^3)c = 0 \)
\( \implies a^3 - k^2c^3 = 0 \) or \( c = 0 \)
Now, if \( a^3 - k^2c^3 = 0 \)
\( \implies k^2 = \frac{a^3}{c^3} \implies (k^2)^{1/3} = \left(\frac{a^3}{c^3}\right)^{1/3} \)
\( \implies k^{2/3} = \frac{a}{c} \)
This is impossible as \( k^{2/3} \) is irrational and \( \frac{a}{c} \) is rational.
\( \therefore a^3 - k^2c^3 \neq 0 \)
From other condition \( c = 0 \).
Substituting \( c = 0 \) in \( b^2 - ac = 0 \), we get \( b = 0 \)
Substituting \( b = 0 \) and \( c = 0 \) in \( a + bk^{1/3} + ck^{2/3} = 0 \), we get \( a = 0 \)
Hence, \( a = b = c = 0 \)

 

 

 

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