<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] First, show that this is true for $n=1$:
$1^3+11=6\cdot2$
Second, assume that this is true for $n$:
$n^3+11n=6k$
Third, prove that this is true for $n+1$:
$(n+1)^3+11(n+1)=$
$n^3+3n^2+14n+12=$
$\color{red}{n^3+11n}+3n^2+3n+12=$
$\color{red}{6k}+3n^2+3n+12=$
$6k+3n(n+1)+12=$


*

*$2    |n\implies6|3n \implies6|3n(n+1)\implies3n(n+1)=6m$

*$2\not|n\implies2|n+1\implies6|3n(n+1)\implies3n(n+1)=6m$


$6k+6m+12=$
$6(k+m+2)$
Please note that the assumption is used only in the part marked red.

Alternatively, consider the following options:


*

*$n\equiv0\pmod6 \implies n^3+11n\equiv  0+ 0\equiv6\cdot 0\equiv0\pmod6$

*$n\equiv1\pmod6 \implies n^3+11n\equiv  1+11\equiv6\cdot 2\equiv0\pmod6$

*$n\equiv2\pmod6 \implies n^3+11n\equiv  8+22\equiv6\cdot 5\equiv0\pmod6$

*$n\equiv3\pmod6 \implies n^3+11n\equiv 27+33\equiv6\cdot10\equiv0\pmod6$

*$n\equiv4\pmod6 \implies n^3+11n\equiv 64+44\equiv6\cdot18\equiv0\pmod6$

*$n\equiv5\pmod6 \implies n^3+11n\equiv125+55\equiv6\cdot30\equiv0\pmod6$

</s>