Question

Which of the following numbers are not perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1728

Answer 1

(i)
On factorising 64 into prime factors, we get:
$64=2\times 2\times 2\times 2\times 2\times 2$

On grouping the factors in triples of equal factors, we get:
$64=\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}$
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.

(ii)
On factorising 216 into prime factors, we get:
$216=2\times 2\times 2\times 3\times 3\times 3$

On grouping the factors in triples of equal factors, we get:
$216=\left\{2\times 2\times 2\right\}\times \left\{3\times 3\times 3\right\}$

It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.

(iii)
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:
$243=\left\{3\times 3\times 3\right\}\times 3\times 3$

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 a not perfect cube.

(iv)
On factorising 1728 into prime factors, we get:
$1728=2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3$

On grouping the factors in triples of equal factors, we get:
$1728=\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{3\times 3\times 3\right\}$

It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.

Thus, (iii) 243 is the required number, which is not a perfect cube.