A simple fact to remember is that the "modulo k" operator can be copied "inside" addition, subtraction and multiplication without altering the result... to say it more formally
$(a + b)_{mod\ k} = (a_{mod\ k} + b_{mod\ k})_{mod\ k}$ $(a - b)_{mod\ k} = (a_{mod\ k} - b_{mod\ k})_{mod\ k}$ $(ab)_{mod\ k} = ((a_{mod\ k})(b_{mod\ k}))_{mod\ k}$ $(a^n)_{mod\ k} = ((a_{mod\ k})^n)_{mod\ k}$
Your problem is to prove that $(n^3 + 6n^2 + 11n + 6)_{mod\ 3} = 0$. Just expanding the problem statement using the above equations you get:
$((n_{mod\ 3})^3 + (6_{mod\ 3})(n_{mod\ 3})^2 + (11_{mod\ 3})(n_{mod\ 3}) + 6_{mod\ 3})_{mod\ 3} = 0$
Observing that $6_{mod\ 3}=0$ and $11_{mod\ 3}=2$ the expression can be simplified to
$ ((n_{mod\ 3})^3 + 2(n_{mod\ 3}))_{mod\ 3} = 0 $
Given that $n_{mod\ 3}$ is either 0, 1 or 2 you only need to test these three cases and this is trivial as it amounts to checking that $(0+0)$, $(1+2)$ and $(8+4)$ are all divisible by 3.
Note of course that it was also quite obvious without any computation that adding or subtracting $6$ or $6n^2$ (both multiples of 3) was not going to change the fact that an expression is divisible by 3 or not.