Prove that the difference of two consecutive cubes is never divisible by 3

Take two consecutive integers:
n, n + 1

The difference of their cubes is:
(n + 1)^3 - n^3
n^3 + 3n^2 + 3n + 1 - n^3

Cancel the n^3
3n^2 + 3n + 1

Factor out a 3 from the first 2 terms:
3(n^2 + n) + 1

The first two terms are always divisible by 3 but then the + 1 makes this expression not divisible by 3:
3(n^2 + n) + 1 = 1 (mod 3)