Problems based on Numbers Test 3

The given equation is ${\mathit{n}}^3-\mathit{n}$.
$=\mathit{n}\left({\mathit{n}}^2-1\right)$
If n is odd, then $\left({\mathit{n}}^2+1\right)$ will be even.
If n is even, then $\left({\mathit{n}}^2-1\right)$ will be odd.
The product of an even and an odd number is always divisible by 6.
Hence, ${\mathit{n}}^3-\mathit{n}$ will be always divisible by 6.