Prove an integer

\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6} will be an integer if \frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6} - n = \left(\frac{n}{3}-\frac{n}{3}\right ) + \left(\frac{n^2}{2} - \frac{n}{2} \right) + \left(\frac{n^3}{6}- \frac{n}{6} \right) is an integer

Since n(n-1) is divisible by 2 and n³-n = (n-1)(n)(n+1) is divisible by 6, the result is true.

This is a more general method I feel, because I have seen such qns with higher powers where factorisation will not be an easy option

@theprophet
i agree yours is a more general method