Let [tex]m,n[/tex] be any two integers, and assume [tex]m-n[/tex] is even. (This would mean either both [tex]m,n[/tex] are even or odd, but that's not important.)
We have
[tex]m^3-n^3=(m-n)(m^2+mn+n^2)[/tex]
and the parity of [tex]m-n[/tex] tells us [tex]m^3-n^3[/tex] must also be even. QED
