= Solution
Factor the expression as $n^3-n=(n-1)n(n+1)$. One of three consecutive <integers> is divisible by $3$, so $3\mid n^3-n$. If $n$ is odd, its neighbours $n-1$ and $n+1$ are consecutive even <integers>. One is divisible by $4$ and the other by $2$, so their product is divisible by $8$. Since $3$ and $8$ are <coprime integers>, their product divides the expression:
$$
\boxed{24\mid n^3-n\quad\text{for every odd integer }n.}
$$
The argument also includes negative odd <integers> and the zero product at $n=\pm1$.
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