Having $n$ colors, use the lemma to find a formula for the number of ways to color the edges of the cube.
Here is what I have so far: The Burnside lemma says that $\displaystyle |X/G| = \dfrac{1}{|G|} \sum_{g \in G} |X^g|$ where $G$ is a finite group, $X^g$ is the set of elements fixed by $g \in G$, and $|x/G|$ is the number of orbits. Now, using $n$ colors, we have the following:
1 identity element leaving $n^6$ elements of $X$ unchanged.
6 90-degree face rotations each leaving $n^3$ elements unchanged.
3 180-degree face rotations each leaving $n^4$ elements unchanged.
8 120-degree vertex rotations each leaving $n^2$ elements unchanged.
6 180-degree edge rotations each leaving $n^3$ elements unchanged.
Using the above results and the formula from Burnside's lemma, we obtain $\dfrac{n^6+6 \cdot n^3 + 3 \cdot n^4 + 8 \cdot n^2 + 6 \cdot n^3}{24} = \dfrac{n^6 + 3n^4 + 12n^3 + 8n^2}{24}$.
I think I did it right but wanted to check with you guys. Thank you!
The algorithm is correct. But note that you are coloring edges. What you did is not.
For example, for identity, there are actually $n^{12}$ elements fixed. Because there are $12$ edges and each edge has $n$ colors.