<s>[INST] Proving $n^3 + 3n^2 +2n$ is divisible by $6$ The full question is: Factorise $n^3 + 3n^2 + 2n$. Hence prove that when $n$ is a positive integer, $n^3 + 3n^2 + 2n$ is always divisible by $6$.
So i factorised and got $n(n+1)(n+2)$ which i think is right? I'm not sure how to actually prove this is divisible by $6$ though. Thanks for help and i apologise if someone has already asked this, i couldn't find it. Also i've not been told whether i have to do induction or not. 
 [/INST] Keep going with your idea. $n(n+1)$ is the product of two consecutive integer so one of them is even. If $n$ is even we are done and if not $n+1$ is and the product is therefore divisible by $2$.
Similarly $n(n+1)(n+2)$ is the product of three consecutive integers so one of them is divisible by $3$. Let's test the residues modulo $3$
$$\begin{array}{c | c c c}
n & n+1 & n+2 & n(n+1)(n+2)\\
\hline
0 & 1 & 2 & 0\\
1 & 2 & 0 & 0\\
2 & 0 & 1 & 0
\end{array}$$
So the product is also divisible by $3$ and therefore the product $n(n+1)(n+2)\equiv 0\pmod 6$
One could also test directly the congruences modulo $6$
$$\begin{array}{c|c c c c}
n & n^3 & 3n^2 & 2n & n^3+3n^2+2n\\
\hline
0 & 0 & 0 & 0 & 0\\
1 & 1 & 3 & 2 & 0\\
2 & 2 & 0 & 4 & 0\\
3 & 3 & 3 & 0 & 0\\
4 & 4 & 0 & 2 & 0\\
5 & 5 & 3 & 4 & 0
\end{array}$$
And we have proven that $n^3+3n^2+2n\equiv 0\pmod 6$
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