Answer
See below.
Work Step by Step
1. Let $P(n)$ be the statement to be proved.
2. For $n=0$, we have $0^3-0=0$ which is divisible by $6$, thus $P(0)$ is true.
3. Assume $P(k), k\gt0$ is true, that is $k^3-k$ is divisible by $6$.
4. For $n=k+1$, we have $(k+1)^3-(k+1)=k^3+3k^2+3k+1-k-1=(k^3-k)+3k(k+1)$
which is divisible by $6$ because $k(k+1)$ is even.
5. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.