Cubic of Integer

The Problem

Prove that the cube of any integer has one of the following forms modulo 9:

$n^3\equiv 0,1,\text{ or }8(\mathrm{mod}9).$

Equivalently, the only possible cubes modulo 9 are $\{0,1,8\}$ (which equals $\{0,1,-1\}$ mod 9).

Use this to show: if $n$ is a positive integer such that the sum of its digits (in base 10) is divisible by 9, then $n^3$ is divisible by 27.