Answer
Let $P(n)$ denote the proposition: "6 divides $n^3 -n$ whenever $n$ is a nonnegative integer."
$P(1)$ is true, since $(1)^3 -(1) = 0$.
Assume that $P(k)$ is true for an arbitrary, nonnegative integer $k$. Then
\begin{align}
(k+1)^3 - (k+1) &= k^3 + 3k^2 +3k +1 -k -1 \\
&= [k^3 -k] + 3[k^2 +k]
\end{align}
The first bracketed term is divisible by 6, via the inductive hypothesis; the second term is divisible by both 3 and 2 (via exercise 31), so $P(k+1)$ is also true. Therefore by method of mathematical induction $P(n)$ is true - 6 divides $n^3 -n$ whenever $n$ is a nonnegative integer.
Work Step by Step
First, we must give the proposition. We let $P(n)$ denote the proposition: "$6$ divides $n^3 - n$ whenever $n$ is a nonnegative integer."
Next, we must prove our base case $P(1)$ is true by showing that the proposition is true when $n=1$. When $n=1$, $n^3 -n = (1)^3 -(1) =0$. $0$ is divisible by $6$, so therefore $P(1)$ is true.
Finally, we must carry out the inductive step, Assume that $P(k)$ is true; that is, we assume that $k^3 -k$ is divisible by $6$ for an arbitrary, nonnegative integer $k$. Now we must show that this implies that $P(k+1)$ is also true. A useful technique to remember in this proof is the method of binomial expansion using Pascal's triangle.
\begin{align}
(k+1)^3 - (k+1) &= {3 \choose 0}k^3 +{3 \choose 1} k^2 + {3 \choose 2} k + {3 \choose 3} - k - 1 \\
&= k^3 + 3k^2 +3k +1 -k -1 \\
&= k^3 -k + 3k^2 +3k \\
&= [k^3 -k] + 3[k^2 +k]
\end{align}
The first bracketed term is divisible by $5$, via our inductive hypothesis. Referring to excercise 31 (which showed that $2$ divides $n^2 +n$), we know that the term $3[k^2 +k]$ is divisible by both $2$ and $3$. By extension $3[k^2 +k]$ must also be divisible by $6$, which means that the entire expression must also be divisible by $6$, and that $P(k+1)$ is true. Therefore by method of mathematical induction $P(n)$ is true - $6$ divides $n^3 -n$ whenever $n$ is a nonnegative integer.
