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: $: $3$ is a factor of $1(1+1)(1-1)=0$
2) Assume for $n=k: 3$ is a factor of $k(k+1)(k-1)=k^3 - k$.
Then for $n=k+1$: $(k+1)((k+1)+1)((k+1)-1)=k^3 + 3 k^2 + 2 k=(k^3-k)+(3k^2+3k)=(k^3-k)+3(k^2+k)$.
The first term of the sum of is divisible by $3$ by the inductive hypothesis and the second one is divisible by $3$ because of the factor, thus their sum is also divisible by $3$.
Thus we proved what we wanted to.