Question

Which of the following numbers are not perfect cubes?
$\left(ii\right)216$

Answer 1

Check if $216$ is a perfect cube

Prime factors of $216$:

$216=2\times 2\times 2\times 3\times 3\times 3$

On grouping,we get
$216=\{2\times 2\times 2\}\times \{3\times 3\times 3\}$

Clearly, the prime factors of $216$ can be grouped in triplets of equal factors and no factor is left over.

Hence, $216$ is a perfect cube.