# Exercise 6.3 class 8 solution

#### Exercise 6.3 class 8 solution-Introduction

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#### Exercise 6.3 class 8 solution- Find smallest multiple

In last section we have observed that every number is not a perfect cube. If a number is not a perfect cube, we can find the smallest natural number by which the given number must be multiplied, so that the product is a perfect cube. We can also find the smallest number to divide the given number, so that quotient is a perfect cube.

#### Exercise 6.3 class 8 solution -examples of the exercise

Example 6.3.1. Is 243 a perfect cube? If not,find the smallest number by which 243 must be multiplied so that the product is a perfect cube. Also find the number.

Solve.

The prime factorization of 243 is 3x3x3x3 × 3 = 3^3 × 3^2

The prime factor 3 does not appear in a group of three. So 243 is not a perfect cube. To make it perfect cube, we need one more 3, so to make 243 a perfect cube. we have to multiply it by 3.

243* 3=729

3х3х3х3х3х3=729

Now, it is a perfect cube number. Hence, the smallest number to be multiplied is 3 and 729 is the number which is perfect cube.

Example 6.3.2 Is 675 a perfect cube? If not, find the smallest natural number by which 675 must be multiplied so that the product is a perfect cube.

Example 6.3.3 Is 31944 a perfect cube? If not, then by which smallest natural number should 31944 be divided so that quotient is a perfect cube?

Solve.

Hence the smallest number by which 31944 should be divided to make it a perfect cube is 3.

#### Exercise 6.3 class 8 solution- exercise preview

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#### Exercise 6.3 class 8 solution- solution pdf

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