Question

"Question 2 Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100

Class 8 Cubes and Cube Roots Page 114"

Answer 1

In the prime factorization of a perfect cube every  prime factor occurs 3 times.

 

To determine whether a number is a perfect cube or not proceed as follows:

1.Find the prime factors of the given number.

2. Make Group of 3 equal prime factors.

3. If a group contains only one or two equal prime factors then a given number is not a perfect cube otherwise it is a perfect cube.

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Solution:

 (i) Prime factors of 243 = (3 × 3 × 3) × 3 × 3

 

Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required.so, we multiply 243 by 3 to make it a perfect cube.

 

243 × 3 = (3 × 3 × 3) × (3 × 3 × 3) = 729 is a perfect cube.

Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

 

(ii) Prime factors of  256 = (2 × 2 × 2) ×( 2 × 2 × 2) × 2 × 2

 

Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required. so, we multiply 256 by 2 to make it a perfect cube

 

256 × 2 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = 512 is a perfect cube.

 

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

 

(iii) Prime factors of 72 = (2 × 2 × 2) × 3 × 3

 

Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required. so, we multiply 72 by 3 to make it a perfect cube.

 

72 × 3 = (2 × 2 × 2) × (3 × 3 × 3) = 216 is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

 

(iv)  Prime factors of 675 = (3 × 3 × 3) × 5 × 5

Here, two 5s are left which are not in a triplet. To make 675 a cube, one more 5 is required. so, we multiply 675 by 5 to make it a perfect cube.

 

675 × 5 = (3 × 3 × 3 )× (5 × 5 × 5 )= 3375 is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

 

(v) Prime factors of 100 = (2 × 2) × (5 × 5)

Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5. so, we multiply 100 by (2 x5 =10) to make it a perfect cube

 

100 × 2 × 5 = (2 × 2 × 2) ×( 5 × 5 × 5)  = 1000 is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10.

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Hope this will help you....

Answer 2

Answer:

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