Prove that if $a$ is an integer, then $a(a^2+2)$ is divisible by $3$ I've been working on this problem for a while.

I've been trying a contrapositive contradiction proof, as the other examples of direct proofs I've seen aren't very clear.

For contrapositive, I assume $3$ does not divide $a(a^2 + 2)$, and $a$ is an integer. Then $a(a^2 + 2) = 3k + 1$ or $a(a^2 + 2) = 3k + 2.$

I'm thinking through simplifying/rearranging $a(a^2 + 2) = 3k + 1$ into $(a^2 + 2a - 2) / 3  = k$ might be sufficient as a contradiction, because since $a$ is an integer, the top portion is also an integer (since integers are closed under addition/subtraction. So an integer divided by $3$ is sometimes not an integer. But I'm not sure if I'm on the right track or not. 

Thank you in advance for any help!
It's pretty easy to prove that if $3$ is not a factor of $a$, it is a factor of $a^2+2$ by noting that $a^2\equiv0,1\bmod3$ (see here)

The rest is simple.
It's pretty easy to prove that if $3$ is not a factor of $a$, it is a factor of $a^2+2$ by noting that $a^2\equiv0,1\bmod3$ (see here)

The rest is simple.