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.

HINT: 
$$\begin{align} & (n^2+n)(n^2+2) \\
&=n(n+1)(n^2+2) \\
&=n(n+1)(n^2-1+3)\\
&=\color{red}{n(n+1)(n-1)} \cdot (n+1)+3\cdot\color{blue}{n(n+1)}\end{align}$$
Get the clue?
