divisibility - Prove that for every natural $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$

Prove that for every natural number $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$.

I've noticed that $(n^2 + n) = n(n+1)$ so these are two successive numbers hence one of them can be divided by two.

I suppose that I should prove that $(n^2 + n)(n^2 + 2)$ can be divided by $3$ but I don't know how to do that.

Answer

Either $n$ is divisible by $3$, or $n^2+2=(n+1)(n-1)+3$ is.