Answer
See below.
Work Step by Step
Proofs using mathematical induction consist 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 we have:
1) For $n=1: 1^3-1+3=3$ is divisible by $3$.
2) Assume for $n=k: k^3-k+3$ is divisible by $3$. Then for $n=k+1$:
$(k+1)^3-(k+1)+3=k^3+3k^2+2k+3=(k^3-k+3)+3(k^2+k)$. The first part of the sum is divisible by $3$ by the induction hypothesis, whilst the second part is divisible because of the factor. Hence the sum is also divisible.
Thus we proved what we wanted to.