<s>[INST] 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
 [/INST] $$3k^2 + 3k + 12=3(k^2 + k +4)= 3(k(k+1)+4)$$
Can you see why $k(k+1)$ and $4$ are each divisible by $2$?
At least one of $k, k+1$ is even, as is $4$, hence $2$ divides $(k(k+1)+4)$, and with three as a factor of $\color{blue}{3}(k(k+1)+ 4)$, we have  $$2\cdot 3 = 6\mid (3k^2 + 3k + 12).$$
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