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.

Here is a solution using induction:
Let $f(x)=x^3+6x^2+11x+6$
Since we want to see if it is divisible by 3 let us assume that $f(x)=3m$.
For the case where $x=0$, $f(0)=6$ which is divisible by 3. 
Now that we have proved for one case let us prove for the case of $f(x+1)$
$$f(x+1)=(x+1)^3+6(x+1)^2+11(x+1)+6$$
$$= x^3+3x^2+3x+1+6x^2+12x+6+11x+11+6$$
$$=(x^3+6x^2+11x+6)+3x^2+15x+18$$
And since $x^3+6x^2+11x+6=3m$
$$f(x+1)=3m+3x^2+15x+18=3(m+x^2+5x+6)$$
Which is divisible by 3.
