What is the number of colourings for cube edges in 5 Blue, 3 Green, 4 Red?
Using Burnside's lemma, I've calculated the group cycle index, which is:
$$ P(G) = \frac{1}{24}(a^{12}+8a^{4}+6a^{3}+3a^{6}+6a^{7})$$
But what's the next step? Any help would be appreciated!
For this computation we need the cycle index $Z(E)$ of $E$, the action of the rotations of the cube on the edges which we now compute. There is the identity, which contributes $$a_1^{12}.$$ Rotations about an axis passing through a pairs of opposite vertices contribute $$4\times 2 a_3 ^4.$$ Rotations about an axis passing through opposite faces contribute $$3 \times (2 a_4^3 + a_2^6).$$ Finally rotations about an axis passing through midpoints of opposite edges contribute $$6\times a_1^2 a_2^5.$$
This gives the cycle index $$Z(E) = \frac{1}{24} \left(a_1^{12} + 8 a_3^4 + 6 a_4^3 + 3 a_2^6 + 6 a_1^2 a_2^5\right).$$
Let's verify this cycle index before we proceed. For edge colorings with $n$ colors where the colors are not being permuted we obtain the formula $$\frac{1}{24} \left(n^{12} + 8 n^4 + 6 n^3 + 3 n^6 + 6 n^7\right).$$ This gives the sequence $$1, 218, 22815, 703760, 10194250, 90775566, 576941778, 2863870080,\ldots$$
which points us to OEIS A060530, where we find that indeed we have the right cycle index.
The desired quantity is by the Polya Enumeration Theorem,
$$[R^4 G^3 B^5] Z(E; R+G+B) \\ = [R^4 G^3 B^5] \frac{1}{24} \left((R+G+B)^{12} + 8 (R^3 + G^3 + B^3)^4 + 6 (R^4 + G^4 + B^4)^3 \\ + 3 (R^2 + G^2 + B^2)^6 + 6 (R + G + B)^2 (R^2 + G^2 + B^2)^5\right) \\ = [R^4 G^3 B^5] \frac{1}{24} (R+G+B)^{12} + [R^4 G^3 B^5] \frac{1}{4} (R + G + B)^2 (R^2 + G^2 + B^2)^5 \\ = \frac{1}{24} {12\choose 4, 3, 5} + \frac{1}{2} [R^4 G^2 B^4] (R^2 + G^2 + B^2)^5 \\ = \frac{1}{24} {12\choose 4, 3, 5} + \frac{1}{2} [R^2 G^1 B^2] (R + G + B)^5 \\ = \frac{1}{24} {12\choose 4, 3, 5} + \frac{1}{2} {5\choose 2, 1, 2} = 1170.$$
Some of this material appeared at the following MSE link.