In set of $8$ cube (each has edge with length $1$) we have $3$ cubes with exactly one white side (other sides are black) and $5$ cubes with all white sides. We make from this cubes, one big cube with size $2$. Count on how many ways can it be done if two ways of making big cube we consider as the same if one big cube can be transformed into to second cube in some rotation of that cube in $\mathbb R^3$. Assume that walls are opaque.
my try:
Let consider rotation of normal cube $1\times 1 \times 1$
\begin{array}{|c|c|c|c|} \hline rotation& \mbox{how many of that type }&cycles\\ \hline id & 1 & x_1^6 \\ \hline \mbox{center of sides 90 degree} & 1 & x_1^2x_4 \\ \hline \mbox{center of sides 180 degree} & 6 & x_1^2x_2^2 \\ \hline \mbox{diagonals} & 3 & x_3^2 \\ \hline \mbox{center of edges} & 8 & x_2^3 \\ \hline \end{array}
so $$I(x_1,x_2,x_3,x_4,x_5,x_6) = \frac{1}{24}(x_1^6+x_1^2x_4+6x_1^2x_2^2+3x_3^2+8x_2^3) $$
but due to we have cube $2\times 2 \times 2$ we rotate $4$ sides per one old side. So I should multiply each cycle by $4$. And get: $$I(x_1,x_2,x_3,x_4,x_5,x_6) = \frac{1}{24}(x_1^{24}+x_1^4x_4^4+6x_1^8x_2^8+3x_3^8+8x_2^{12}) $$ but I am not sure what have I to do now?
There are 3 ways to position the cubes painted white on only one side.
They can all be on the same side. Two can be adjacent, and one non adjacent, or all 3 can be non adjacent.
For each partially white cube, there 3 orientations where the white face points out. But, it could be hidden, too.
The rightmost has a symmetry to it that must be accounted for separately.
there are 4 configurations that will be invariant under rotation. For the others we need to divide by 3.
$4^3 + 4^3 + 4 + \frac {4^3-4}{3}\\ 64+ 64 + 4 + 20 = 152$