$\displaystyle 2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$
$\displaystyle 6|(2n^3+3n^2+n)\rightarrow 6k=2n^3+3n^2+n$
???
Hello, dwsmith!
I have a back-door approach to this one . . .
$\displaystyle 2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$
As aman-cc pointed out: .$\displaystyle 2n^3 +3n^2 + n \:=\:n(n+1)(2n+1)$
I recognized this is as part of the formula for the sum of consecutive squares:
. . $\displaystyle 1^2 + 2^2 + 3^2 + \hdots + n^2 \:=\:\dfrac{n(n+1)(2n+1)}{6}$
Since the left side is an integer, the right side must be an integer.
. . That is, $\displaystyle n(n+1)(2n+1)$ is a multiple of 6.
Therefore: .$\displaystyle 2n^3 + 3n^2 + n \:\equiv\:0\text{ (mod 6)}$ .for all integer values of $\displaystyle n.$
$\displaystyle 2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$
This a solution I saw once:
$\displaystyle 2n^3+3n^2+n=n(2n^2+3n+1)=n(n+1)(2n+1) $. One of $\displaystyle n $ and $\displaystyle n+1 $ is divisible by $\displaystyle 2 $.
$\displaystyle 2\equiv -1(mod\ 3)$;
$\displaystyle n(n+1)(2n+1)\equiv -(n-1)n(n+1)(mod\ 3)$. Then at least one of $\displaystyle n-1 $, $\displaystyle n $, and $\displaystyle n+1 $ is divisible by $\displaystyle 3 $.
Since $\displaystyle 2 $ and $\displaystyle 3 $ are relatively prime $\displaystyle 2n^3+3n^2+n $ is divisible by $\displaystyle 6 $.