Q:

If a is an arbitrary integer, then $6|a(a^2+11)$

Pprove: If a is an arbitrary integer, then $6|a(a^2+11)$ 
So I substituted $k+1$ in for a and for my final step I have: 
$6|k^3+3k^2+14k+12$ 
At this point what I have left isn't divisible by $6$ so I'm not sure where to go from here.
Step by step explanation please!

A:

Since $6|12$ and $6|a(a+1)(a-1)$, $6|a(a^2-1),$ (product of three sequential numbers is divisible by 6), we have that $6|(12 a+a(a+1)(a-1))$, $6|a(a^2+11)$.

