Non equivalent colourings of regular hexagon( Brualdi Chapter-14 , Exercise -32) I have a question in this exercise of Richard Brualdi's Introductory Combinatorics. 
Exercise is -> Determine the number of non equivalent colourings of corners of regular hexagon with colours red, white and blue. 
Now, Taking motivation from this example solved in text
Adding image of example->
What I calculated  for regular hexagon $ N(D_6, C) $ = $\frac { 3^6 + 5×3 + 6 × 3^3} {12}$=75.5 . 
It seems I am making some mistake as non equivalent colouring comes out to be fractional. I tried to solve it again, but I am getting same answers. 

Can some please tell what I am doing wrong. 


The dihedral group of the hexagon is $\rho^0,\rho^1,\rho^2,\rho^3,\rho^4,\rho^5,\tau_1,\tau_2,\tau_3,\sigma_4,\sigma_5,\sigma_6$.
The $\rho^i$ are the rotations, the $\tau_i$ are the reflections through axes which pass though the vertices of the hexagon and the $\sigma_i$ are the reflections which do not pass through the vertices of the hexagon.
For each permutation $g$, the number of cycles $c(g)$ and $3^{c(g)}$ are listed below:
$$\rho^0 \quad 6\quad  729\\ \rho^1\quad 1 \quad 3 \\ \rho^2\quad 2 \quad 9\\ \rho^3 \quad 3\quad  27 
\\ \rho^4 \quad2 \quad 9\\ \rho^5\quad 1 \quad 3 \\ \sigma_i \quad 3 \quad 27\\ \tau_i\quad 4\quad 81$$
The number of nonequivalent colourings is $\frac{1}{|G|}\sum 3^{c(g)}=\frac{1104}{12}=92$.
