Properly Painted Cubes

Worked solution

The answer is 55. First count proper colourings of a fixed cube. Faces correspond to vertices of an octahedron, so we need the octahedron's chromatic polynomial. Choose colours for the three mutually adjacent faces top, front, right ($5 \cdot 4 \cdot 3 = 60$ ways), then count ways to complete; summing cases (each opposite face either repeats its partner's colour or uses a fourth or fifth colour) gives 13 completions, so 780 proper colourings. Now Burnside over the 24 rotations. Any rotation that puts two adjacent faces in a common cycle fixes nothing: that rules out the quarter-turns (the four side faces form one cycle), the vertex rotations (three faces at a corner are mutually adjacent) and the edge half-turns (they swap the two faces meeting at the axis edge). A face half-turn fixes top and bottom and swaps front with back and left with right; fixed proper colourings choose front = back $= a$, left = right $= b \neq a$, then top and bottom independently from the 3 remaining colours: $5 \cdot 4 \cdot 3 \cdot 3 = 180$. So the count is $(780 + 3 \cdot 180)/24 = 1320/24 = 55$.