#### Answer

See below.

#### Work Step by Step

Proofs using mathematical induction consists of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$.
Hence here:
1) For $n=1: 1^3+2(1)=3$ and this is divisible by $3$.
2) Assume for $n=k: k^3+2k$ is divisible by $3$. Then for $n=k+1:(k+1)^3+2(k+1)=k^3+3k^2+3k+1+2k+2=k^3+3k^2+5k+3$ and we know that $k^3+2k$ is divisible by $3$ by the inductive hypothesis, $3k^2+3=3(k^2+1)$ is also divisible by $3$, thus their sum is also divisible by $3$. Thus we proved what we wanted to.