For each of the non-perfect cubes in Q. No. 20 find the smallest number by which it must be

The only non-perfect cube in question number 20 is 243.

(a)
On factorising 243 into prime factors, we get:

$243=3 \times 3 \times 3 \times 3 \times 3$

On grouping the factors in triples of equal factors, we get:

$243=\{3 \times 3 \times 3\} \times 3 \times 3$

It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is multiplied by 3, the factors can be grouped into triples of equal factors such that no factor is left over.

Thus, 243 should be multiplied by 3 to make it a perfect cube.

(b)
On factorising 243 into prime factors, we get:

$243=3 \times 3 \times 3 \times 3 \times 3$

On grouping the factors in triples of equal factors, we get:

$243=\{3 \times 3 \times 3\} \times 3 \times 3$

It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is divided by $(3 \times 3=9)$, the factors can be grouped into triples of equal factors such that no factor is left over.

Thus, 243 should be divided by 9 to make it a perfect cube.