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# Find the smallest number by which 243 must be

**(a)** multiplied so that the product is a perfect cube.

**(b)** divided so that the quotient is a perfect cube.

To do:

We have to find the smallest number by which 243 must be

(a) multiplied so that the product is a perfect cube.

(b) divided so that the quotient is a perfect cube.

Solution:

Prime factorisation of 243 is,

$243=3\times3\times3\times3\times3$

$=3^3\times3^2$

Grouping the factors in triplets of equal factors, we see that two factors $3 \times 3$ are left.

(a) In order to make 243 a perfect cube, we have to multiply it by 3.

$243\times3=3\times3\times3\times3\times3\times3$

$=3^3\times3^3$

The smallest number by which 243 must be multiplied so that the product is a perfect cube is 3.

(b) Therefore, dividing $243$ by $3\times3=9$, we get,

$243\div9=3\times3\times3\times3\times3\div9$

$=3\times3\times3$

Hence, the smallest number by which the given number must be divided so that the quotient is a perfect cube is 9.

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