Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$. Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6. 
$P(k+1)=(k+1)^3+11(k+1)=k^3+3k^2+3k+1+11k+11=(k^3+11k)+(3k^2+3k+12)$
Since $P(k)$ is true, $(k^3+11k)$ is divisible by 6 but I can't show that $(3k^2+3k+12)$ is divisible by 6

Since
$n^{3} + 11n = 6m$
for some integer $m,$
we have
$$(n+1)^{3} + 11(n+1) = 6m + 3n^{2} + 3n + 12 = 6m + 12 + 3(n^{2} + n).$$
It suffices to prove that $3(n^{2} + n)$ is a multiple of $6$. But, since if $n$ is odd then $n^{2} + n = 2m'$ for some integer $m'$ and if $n$ is even then of course $n^{2} + n = 2m''$ for some integer $m'',$ it follows that $6$ is indeed a multiple of $3(n^{2}+n),$ qed.
