Paint a Cube with Three Colors

Problem: Given a cube, each face can be painted with one of the three colors. Calculate the ways the cube can be colored.

Solution: The problem can be solved by enumeration. While a more formal way is to use Burside's Lemma. Generally we need to understand the symmetry of the permutation on cubes. The main ideas of this method are listed as follows:

1. find the permutation groups of a cube, the easiest one is the identity permutation (just don't change). Then for the six faces, we have 1->1, 2->2 ... and 6->6. So for this permutation, there are 6 circles and the length of each circle is 1. Therefore we have a1^6, where a1 means the sub group that with circle length 1. "6" means there are 6 such circles.
2. then we can rotate along the axis that penetrates the centers of two opposite faces for 90 degree. If we do this, we have a1^2*a4, since two face unchanged, and the rest 4 form a circle. We have 6 such rotations. Therefore at the end we have 6*a1^2*a4.
3. then we can rotate along the axis that penetrates the centers of two opposite faces for 180 degree. If we do this, we have a1^2*a2^2.We have 3 such rotations. Therefore at the end we have 3*a1^2*a2^2.
4. then we can rotate the two opposite vertices (diagonal). We will have a3^2, since three faces that share one vertex are in one circle. We have 8 such rotations. Therefore at the end we have 8*a3^2.
5. then we can rotate along the axis that penetrates the middle points of two parallel edges that are not in the same face for 180 degree. We will have 6*a2^3 at the end.
6. so totally we have 1+6+3+8+6 = 24ways of permutations. So the cycle index (the ways to paint the cube) = 1/24*(a1^6+6*a1^2*a4+3*a1^2*a2^2+8*a3^2+6*a2^3). Since a1=a2=...=a6, we have cycle index =  1/24*(a^6+3*a^4+12*a^3+8*a^2). 
7. when a = 3, we have cycle index = 57.