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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.