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...