<s>[INST] prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$ I'm having some trouble with this question and can't really get how to prove this..
I have to prove $n^3+6n^2+11n+6$ is divisible by $3$ for all $n \geq 0$. 
I have tried doing $\dfrac{m}{3}=n$ and then did $m=3n$
then I said $3n=n^3+6n^2+11n+6$ but now I am stuck.
 [/INST] We have
$$
\begin{align}
n^3+6n^2+11n+6
&=6\binom{n}{3}+18\binom{n}{2}+18\binom{n}{1}+6\binom{n}{0}\\
&=6\left(\binom{n}{3}+3\binom{n}{2}+3\binom{n}{1}+\binom{n}{0}\right)
\end{align}
$$
so $6\mid(n^3+6n^2+11n+6)$ for all $n\in\mathbb{Z}$.
Of course, since $3\mid 6$, we have $3\mid(n^3+6n^2+11n+6)$, as requested.
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