Pólya Enumeration Theorem and Chemical Compounds How many different chemical compounds can be made by attaching H, CH3, or OH radicals to each of the carbon atoms in the benzene ring of Figure 1? (Assume that all of the C–C bonds in the ring are equivalent.)
Figure 1:

I understand the overall method of the Pólya-Burnside Method of Enumeration but I don't understand how to apply it to a chemical compound problem like this. Any help would be great, thank you in advance!

Here is a solution via the Polya Enumeration Theorem.  The problem is the same as determining the number of ways of coloring the vertices of a hexagon with three colors.
The symmetry group of the hexagon is the dihedral group $D_6$.  The cycle index of the group is
$$Z=\frac{1}{12} (x_1^6 + 4 x_2^3 +2 x_3^2 +2 x_6 + 3x_1^2 x_2^2)$$
The figure inventory for three colors (or three radicals) is $x+y+z$. Substituting the figure inventory into the cycle index yields
$$\frac{1}{12}[ (x+y+z)^6 + 4(x^2+y^2+z^2)^3  +2(x^3+y^3+z^3)^2 + 2(x^6+y^6+z^6) + 3(x+y+z)^2(x^2+y^2+z^2)^2  ]$$
If we were to expand this polynomial we could find all the possible ways of coloring the vertices with specific combinations of colors, such as one red vertex, two blue vertices, and three green vertices.  But all we really want to know for this problem is the total number of ways with all possible combinations of three colors, which is the sum of the coefficients, and we can find that sum by setting $x=y=z=1$, with result
$$\frac{1}{12}(3^6 + 4 \cdot 3^3 + 2 \cdot 3^2 + 2 \cdot 3 + 3 \cdot 3^2 \cdot 3^2) = \boxed{92}$$
