Proof of $n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$ by mathematical induction 
Prove the following statement by mathematical induction:
$n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$
My attempt:
Let the given statement be p(n).
(1) $1(1^2+5)$=6 Hence, p(1) is true.
(2) Suppose for all integer $k \ge 1$, p(k) is true.
That is, $k(k^2+5)$ is divisible by 6 
We must show that p(k+1) is true.
$(k+1)((k+1)^2+5)$=$k^3+3k^2+3k+1+5(k+1)$
=$k^3+3k^2+8k+6$
=$k(k^2+5)+3k^2+3k+6$
I'm stuck on this step. I feel I have to show $3k^2+3k+6$ is divisible by 6. But, how can I show $3k^2+3k+6$ is divisible by 6?

Hint:
let $$f(n)=n(n^2+5),n\geq1$$ then
$$f(n+1)=(n+1)(n^2+2n+6)=n^3+2n^2+6n+n^2+2n+6=n^3+3n^2+8n+6=$$
$$=n^3+5n+3n^2+3n+6=3(n^2+n)+6+n(n^2+5)=f(n)+6\left(\frac{n(n+1)}{2}+1\right)$$
