Solution: We need to prove that $k^3 + 2k $ is divisible by $3$. We recall that if $p$ is a prime and $p \mid (ab)$, then $p \mid a$ or $p \mid b$. Note that \begin{align*} 3 \mid (k^3 + 2k) & \iff 3 \mid k (k^2 + 2) \iff 3 \mid k \text{ or } 3 \mid (k^2 + 2). \end{align*} If $3 \mid k$, then we are done. If not, then $ k = 3m + 1$ or $3m + 2$. Then \begin{align*} k = 3m + 1 \implies k^2 + 2 & = (3m + 1)^2 + 2 \\ & = 9m^2 + 6m + 1 + 2 \\ & = 3 \left( 3m^2 + 2m + 1 \right). \\ k = 3m + 2 \implies k^2 + 2 & = (3m + 2)^2 + 2 \\ & = 9m^2 + 12m + 4 + 2 \\ & = 3 \left( 3m^2 ++ 4m + 2 \right) . \end{align*} Therefore, in either case, $3 \mid (k^2 + 2)$. Hence, we proved that $k^3 + 2k$ is divisible by $3$.