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