prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$

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.

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.

There's an algorithm for finding rational roots of a polynomial with integer coefficients, and if you use that to factor this polynomial you get $$ n^3 + 6n^2 + 11n + 6 = (n+1)(n+2)(n+3). $$ For any $n$, one of those three factors is a multiple of $3$.