coloring cube, additional constraint, three colors I have to paint nodes of cube such that opposing nodes has the same color.
We consider identical cubes such that rotatating.
My result is $15$
Is it correct ?
Ok, I 'll add my way to get a result.
$G$ is group of rotating.
$$G = \{id, o_1,o_2,o_3,r_1,..,r_1, ..., r_6, s_1,...,s_6,k_1,....,k_8\}\\
|G|=24$$
Const points with respect to $id$: $3^4 $ 
Const points with respect to $o_i$: $3*3^2 $ 
Const points with respect to $r_i$: $6 * 3$ 
Const points with respect to $s_i$: $6 * 3^3$ 
Const points with respect to $k_i$: $8*3*3$ 
Finally:
$$\frac{1}{24} (3^4 + 3 * 3^2 + 6 * 3 + 6*3^3 +8* 3 * 3) = 15$$
 A: You used Polya counting theory and arrived at $15$, but are not convinced. So let's count the colorings of the four space diagonals again, this time using "brute force".
There are $3$ colorings using just one color.
If we restrict to just two colors we can do it in a $(3,1)$-way or in a $(2,2)$-way. For the $(3,1)$-way we can select the primary color in three ways and the secondary color in two ways. After the colors have been chosen there is just one way to color the diagonals, so that we obtain $6$ colorings of this sort. For the $(2,2)$-way we can select the two colors to be used in three ways. After the colors have been chosen there is just one way to color the diagonals, so that we obtain $3$ colorings of this sort.
If we use all three colors one of them, which can be chosen in $3$ ways, is used twice, and the other two colors are used once. Paint any two diagonals with these two colors and the remaining two with the chosen color – it always looks the same. It follows that we obtain $3$ colorings of this sort.
So it comes to $15$ colorings, as you have found out using the group theoretical approach.
