What is the smallest number by which 2916 be multiplied so that the product is a perfect cube?

Solution:

We have to find the smallest whole number by which the number should be multiplied so as to get a perfect square number

To get a perfect square, each factor of the given number must be paired.

(i) 252

Hence, prime factor 7 does not have its pair. If 7 gets a pair, then the number becomes a perfect square. Therefore, 252 has to be multiplied by 7 to get a perfect square.

So, perfect square is 252 × 7 = 1764

1764 = 2 × 2 × 3 × 3 × 7 × 7

Thus, √1764 = 2 × 3 × 7 = 42

(ii) 180

Hence, prime factor 5 does not have its pair. If 5 gets a pair, then the number becomes a perfect square. Therefore, 180 has to be multiplied by 5 to get a perfect square.

So, perfect square is 180 × 5 = 900

900 = 2 × 2 × 3 × 3 × 5 × 5

Thus, √900 = 2 × 3 × 5 = 30

(iii) 1008

Hence, prime factor 7 does not have its pair. If 7 gets a pair, then the number becomes a perfect square. Therefore, 1008 has to be multiplied by 7 to get a perfect square.

So, perfect square is 1008 × 7 = 7056

7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

Thus, √7056 = 2 × 2 × 3 × 7 = 84

(iv) 2028

Hence, prime factor 3 does not have its pair. If 3 gets a pair, then the number becomes a perfect square. Therefore, 2028 has to be multiplied by 3 to get a perfect square.

So, perfect square is 2028 × 3 = 6084

6084 = 2 × 2 × 13 × 13 × 3 × 3

Thus, √6084 = 2 × 13 × 3 = 78

(v) 1458

Hence, prime factor 2 does not have its pair. If 2 gets a pair, then the number becomes a perfect square. Therefore, 1458 has to be multiplied by 2 to get a perfect square.

So, perfect square is 1458 × 2 = 2916

2916 = 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2

Thus, √2916 = 3 × 3 × 3 × 2 = 54

(vi) 768

Hence, prime factor 3 does not have its pair. If 3 gets a pair, then the number becomes a perfect square. Therefore, 768 has to be multiplied by 3 to get a perfect square.

So, perfect square is 768 × 3 = 2304

2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

Thus, √2304 = 2 × 2 × 2 × 2 × 3 = 48

☛ Check: NCERT Solutions for Class 8 Maths Chapter 6

Video Solution:

NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.3 Question 5

Summary:

For each of the following numbers, (i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768 the smallest whole number by which it should be multiplied so as to get a perfect square number and the square root of the square number so obtained are as follows: (i) 7; √1764 = 42 (ii) 5; √900 = 30 (iii) 7; √7056 = 84 (iv) 3; √6084 = 78 (v) 2; √2916 = 54 and (vi) 3; √2304 = 48

☛ Related Questions:

Extra Questions for Class 8 Maths Chapter 7 Cubes and Cube Roots

Cubes and Cube Roots Class 8 Extra Questions Very Short Answer Type

Question 1. Find the cubes of the following: (a) 12 (b) -6 (c) \(\frac { 2 }{ 3 }\) (d) \(\frac { -5 }{ 6 }\) Solution:

Question 2. Find the cubes of the following: (a) 0.3 (b) 0.8 (c) .001 (d) 2 – 0.3 Sol.

(a) (0.3)3 = 0.3 × 0.3 × 0.3 = 0.027

(b) (0.8)3 = 0.8 × 0.8 × 0.8 = 0.512
(c) (0.001)3 = (0.001) × (0.001) × (0.001) = 0.000000001
(d) (2 – 0.3)3 = (1.7)3 = 1.7 × 1.7 × 1.7 = 4.913

Question 3. Is 135 a perfect cube? Solution: Prime factorisation of 135, is: 135 = 3 × 3 × 3 × 5 We find that on making triplet, the number 5 does not make a group of the triplet. Hence, 135 is not a perfect cube.

Question 4. Find the cube roots of the following: (a) 1728 (b) 3375 Solution:

Question 5. Examine if (i) 200 (ii) 864 are perfect cubes. Solution: (i) 200 = 2 × 2 × 2 × 5 × 5 If we form triplet of equal factors, the number 2 forms a group of three whereas 5 does not do it. Therefore, 200 is not a perfect cube.

(ii) We have 864 = 2 × 2 × 2 × 2 × 2 If we form triplet of equal factors, the number 2 and 3 form a group of three whereas another group of 2’s does not do so. Therefore, 864 is not a perfect cube.

Question 6. Find the smallest number by which 1323 may be multiplied so that the product is a perfect cube. Solution: 1323 = 3 × 3 × 3 × 7 × 7 Since we required one more 7 to make a triplet of 7. Therefore 7 is the smallest number by which 1323 may be multiplied to make it a perfect cube.

Question 7. What is the smallest number by which 2916 should be divided so that the quotient is a perfect cube? Solution: Prime factorisation of 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 Since we required one more 2 to make a triplet Therefore, the required smallest number by which 2916 should be divided to make it a perfect cube is 2 × 2 = 4, i.e., 2916 ÷ 4 = 729 which is a perfect cube.

Question 8. Check whether 1728 is a perfect cube by using prime factorisation. (NCERT Exemplar) Solution: Prime factorisation of 1728 is 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 Since all prime factors can be grouped in triplets.

Therefore, 1728 is a perfect cube.

Question 9. Using prime factorisation, find the cube root of 5832. (NCERT Exemplar) Solution:

Question 10.

Solution: