Use the principle of mathematical induction to prove that $\displaystyle n(n^2 +5)$ is divisible by 6 for all n=$\displaystyle Z^+$...
Lets start at the point assume $\displaystyle n$ and show that$\displaystyle n \implies n+1$
assume that $\displaystyle 6 |n(n^2+5)$
We need to show that 6 divides $\displaystyle (n+1)((n+1)^2+5)$
$\displaystyle (n+1)((n+1)^2+5)=n^3+3n^2+3n+1+5n+5$
Now grouping the terms like this we get
$\displaystyle (n^3+5n) + 6 +3n(n+1)$
The first group is divisible by 6 from the induction hypothsis 6 is obviously divisible by 6.
Notice that for every integer n $\displaystyle n(n+1)$ is even and so it can be written in the form $\displaystyle 2k$ for some integer k.
$\displaystyle 3n(n+1) =3 \cdot 2k=6k$ this is also divisible by 6.