Show that $n^3-n$ is divisible by $6$ using induction

Solution 1:

No, your argument is not quite right (or at least not clear to me). You must show that if $A(n)$ is true, then $A(n+1)$ follows. $A(n)$ here is the statement "$n^3-n$ is divisible by $6$". Assuming $A(n)$ is true, then $$(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=(n^3-n)+3n(n+1)$$ is divisible by $6$ because (1) $n^3-n$ is a multiple of $6$ by assumption, and (2) $3n(n+1)$ is divisible by $6$ because one of $n$ or $n+1$ must be even (this is related to what Alex was pointing out). Therefore $A(n)$ implies $A(n+1)$ and, if $A(n)$ is true for some value of $n$, then all higher integer values of $n$ follow. You correctly showed that $A(0)$ is true.

Solution 2:

No need for induction. $$n^3-n=n(n^2-1)=n(n-1)(n+1)$$ which are three consecutive integers. So one must be divisible by 3.

Check for $n=1$: $1^3-1=0=3\cdot 0$

Assume it's true for $n=k$. If you let $n=k+1$ you get $$\begin{align*} (k+1)^3-(k+1)&=k^3+3k^2+2 \\ &=k^3+3k^2+2k\\ &=3\cdot (k^2+k)+(k^3-k)\end{align*}$$ which is divisible by 3