Show that the difference of two consecutive cubes is never divisible by $3$.

$(n+1)^3 - n^3 = n^3 + 3n^2 + 3n + 1 - n^3 = 3(n^2 + n) + 1 \equiv 1$ (mod $3$)

This can also be understood geometrically.

Arrange $n^3$ little cubes into a large cube of side length $n$, and then remove $(n-1)^3$ of them, leaving three flat sheets of cubes meeting at three edges.

Every cube left is either on the inside of one of the three sheets (and there are equally many of these on each sheet) or at one of the edges but not at the apex (again by symmetry each edge has equally many cubes in it), or it is the apex cube.

Therefore the total number of cubes left must be three times something, plus one cube at the apex.