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?

• $k^3+3k^2+8k+6=k\left(k^2+5\right)+3\left(k^2+k+2\right)$. Notice $k^2+k+2=k(k+1)+2$ is always even for all $k\in\mathbb Z$. Feb 16, 2016 at 10:25
• Does it have to be by induction? It seems to be quicker to show that it is always congruent to 0 modulo 2 as well as modulo 3, and then appeal to Chinese remainders. Feb 16, 2016 at 10:36
• @user236182 I edited it. Feb 16, 2016 at 11:37
• @buzzee Now notice that $3k^2+3k+6=3(k(k+1)+2)$. This number is divisible by $6$ because $k(k+1)+2$ is divisible by $2$ (because $k(k+1)$ is divisible by $2$, because exactly one of $k,k+1$ is divisible by $2$). Feb 16, 2016 at 11:41