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.

If you know what "mod 3" means then argue as follows:
$$n^3 + 6n^2 + 11n + 6 \equiv n^3 - n = (n-1)n(n+1) \equiv 0 \pmod 3 .$$
If you don't, then write this as:
$$ n^3 - n + 12n + 6n^2 + 6 = n(n+1)(n-1) + 3(2n^2 + 4n + 2), $$
and you're left with showing that both terms are divisible by $3$.
Now $n(n+1)(n-1)$ is always a multiple of $3$, because if a number is not a multiple of 3, then either its predecessor or its successor must be.
