Question

Which of the following numbers are not perfect cubes?
$\left(iv\right)1728$

Answer 1

Check if $1728$ is a perfect cube

Prime factors of $1728$:

$1728=2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3$

On grouping the factors of $216$, we get
$1728=\{2\times 2\times 2\}\times \{2\times 2\times 2\}\times \{3\times 3\times 3\}$

Clearly, the prime factors of $1728$ can be grouped in triplets of equal factors and no factor is left over.

Hence, $1728$ is a prefect cube.