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.

We have
$$n^3 + 6n + 11n + 6 = (n+1)(n+2)(n+3).$$
A run of three consecutive integers has an integer in it divisible by 2, so 
$$2|n^3 + 6n + 11n + 6.$$   A run of three consecutive integers has an integer in it divisible by 3, so 
$$3|n^3 + 6n + 11n + 6.$$ 
Since 2 and 3 are relatively prime, the  desired conclusion follows right away.
