Prove that $\forall n \in Z$, $n^3+3n$ is even.
Attempt: I am solving this problem using proof by cases. Case 1 is when $n$ is even, i.e. $n=2b$. This one is easy. However, in case 2 when $n$ is odd, i.e. ($n=2a+1$) I am having difficulties with showing that $n^3+3n$ is even. Namely, after plugging in and expanding I am stuck: $$n^3+3n=(2a+1)^3+3(2a+1)=8a^3+12a^2+6a+1+3a+3=\space...$$ I tried regrouping the terms, but my effors did not amount to anything. I guess I am making a mistake somewhere. Help appreiciated. Thank you.