We have a classical cube with six faces , eight vertices and twelve edges . We want to color faces ,vertices and edges of this cube using three colors. How many non-equivalent coloring are there ?
My work I calculated that $$\frac{1}{24}\bigg[x_1^{26}+6x_1^2x_4^6 +9x_1^2x_2^{12} +8x_1^2x_3^8\bigg]$$ $$\frac{1}{24}\bigg[3^{26}+6(3^8) +9(3^{14}) +8(3^{10})\bigg]=11769712290$$
Is my solution correct ?
What we need to verify here is the cycle index of the action of the rotations of the cube on the faces, vertices and edges simultaneously.
We first get the identity, wich contributes (use $a$ for faces, $b$ for vertices and $c$ for edges):
$$a_1^6 b_1^8 c_1^{12}.$$
There are rotations about an axis passing through opposite faces (three such axes) which fix those faces, giving
$$3\times (2a_1^2 a_4 b_4^2 c_4^3 + a_1^2 a_2^2 b_2^4 c_2^6).$$
There are rotations about axes passing through opposite vertices (four such axes), giving
$$4 \times 2 a_3^2 b_1^2 b_3^2 c_3^4.$$
Finally there are rotations about axes passing through the midpoints of pairs of opposite edges (six such pairs), which contribute
$$6\times a_2^3 b_2^4 c_1^2 c_2^5.$$
Collecting everything we get the cycle index
$$\frac{1}{24} (x_1^{26} + 6 x_1^2 x_4^6 + 3 x_1^2 x_2^{12} + 8 x_1^2 x_3^8 + 6 x_1^2 x_2^{12}) \\ = \frac{1}{24} (x_1^{26} + 6 x_1^2 x_4^6 + 9 x_1^2 x_2^{12} + 8 x_1^2 x_3^8).$$
We have verified the cycle index by OP. We get the sequence for colorings with at most $n$ colors
$$1, 2802752, 105912891117, 187650085502976, 62088173933203125, \ldots$$
This is the polynomial
$$A_n = \frac{1}{24} (n^{26}+9 n^{14} +8 n^{10}+6 n^8).$$
We can also compute $B_n$ giving colorings with exactly $n$ colors using PIE. Here the underlying poset has nodes $Q\subseteq [n]$ with weight $(-1)^{n-|Q|}$ that represent colorings using some subset of the colors in $Q.$ A coloring that uses exactly the set $P\ne\emptyset$ of colors where $P\subset [n]$ is thus included in all nodes such that $P\subseteq Q$ for a total weight of
$$\sum_{P\subseteq Q \subseteq [n]} (-1)^{|Q|} = \sum_{q=0}^{n-|P|} {n-|P|\choose q} (-1)^{n-q-|P|} = 0.$$
On the other hand a coloring using exactly $n$ colors is only included in $Q=[n]$ with weight $(-1)^{n-|[n]|} = 1$ as required. Hence for colorings with exactly $n$ colors we have
$$B_n = \sum_{Q \subseteq [n]} (-1)^{n-|Q|} A_{|Q|} = \sum_{q=0}^n {n\choose q} (-1)^{n-q} A_q.$$
We get the sequence
$$1, 2802750, 105904482864, 187226450755016, 61150982606571900, \ldots$$
We can also compute $B_n$ using Stirling numbers of the second kind from the cycle index and we obtain
$$B_n = \frac{n!}{24} \left({26\brace n} + 9 {14\brace n} + 8 {10\brace n} + 6 {8\brace n}\right).$$
With this representation we see that the sequence is finite. For $n=26$ the formula evaluates to $26!/24$ because all orbits have the same size, containing $24$ colorings.