AMC 12 Mock Exam 2 Problem 17

Group the faces into three pairs of opposite faces. Two faces from different pairs always share an edge, so no color can appear in two different pairs: the three pairs use disjoint sets of colors. Each pair is either painted with a single color (its two faces are not adjacent) or with two different colors.

All three pairs single-colored: the pairs get three distinct colors, in $4\cdot3\cdot2=24$ ways.

Exactly one pair two-colored: choose that pair ($3$ ways) and paint its two faces with different colors ($4\cdot3=12$ ways); the other two pairs get the remaining two colors, one each ($2$ ways). That is $3\cdot12\cdot2=72$.

Two two-colored pairs would need at least $2+2+1=5$ colors, which is impossible.

The total is $24+72=\boxed{96}$.