Let [tex]n=1[/tex]. Then [tex]n^3-n=1^3-1=0[/tex]. By convention, every non-zero integer [tex]n[/tex] divides 0, so [tex]3\vert n^3-n[/tex].
Suppose this relation holds for [tex]n=k[/tex], i.e. [tex]3\vert k^3-k[/tex]. We then hope to show it must also hold for [tex]n=k+1[/tex].
You have
[tex](k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)[/tex]
We assumed that [tex]3\vert k^3-k[/tex], and it's clear that [tex]3\vert 3(k^2+k)[/tex] because [tex]3(k^2+k)[/tex] is a multiple of 3. This means the remainder upon divides [tex](k+1)^3-(k+1)[/tex] must be 0, and therefore the relation holds for [tex]n=k+1[/tex]. This proves the statement.
