To solve the problem of finding the number of distinct ways to color the vertices of a cube with 3 colors such that no two adjacent vertices have the same color (considering cube symmetries), we follow these key observations:

Key Observations

The cube's vertex graph is bipartite, split into two disjoint sets of 4 vertices each (let’s call them Partition A and Partition B). Adjacent vertices lie in opposite partitions, so no two vertices in the same partition are adjacent.

For a proper coloring (adjacent vertices different),:

• All vertices in Partition A can be colored freely (no adjacency), but their colors must not conflict with adjacent vertices in Partition B.
• However, using 3 distinct colors in a proper coloring is impossible (since each vertex in Partition B would need a color not present in its neighbors, leading to no valid options).

Thus, valid colorings use at most 2 colors (one for each partition).

Distinct Colorings Under Symmetries

Under cube rotations/reflections, the distinct colorings are determined by choosing 2 distinct colors from the 3 available. Each choice of 2 colors corresponds to one unique coloring (swapping the two colors for the partitions is equivalent under symmetry).

The number of ways to choose 2 colors from 3 is $\binom{3}{2}=3$.

Answer

$\boxed{3}$

Final Answer: $\boxed{3}$

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