Pólya-Burnside Method of Enumeration with Necklace

Use the Pólya-Burnside Method of Enumeration or just the Burnside theorem to find the number of different types of circular necklaces that could be made from the following beads (assuming all are used on the necklace):

1: Three black and three white beads.

2: Four black, three white, and one red.

In these problems, the necklace composition is considered equivalent to another if you can flip the necklace over or rotate it such that it matches the other.

I don't know how to do this. I understand the method of enumeration but I don't know how to apply it to here. Any help would be great, thank you in advance!

1. The cycle index polynomial of $$D_6$$ is $$\frac{1}{12}[x_1^6+3x_1^2x_2^2+4x_2^3+2x_3^2+2x_6]$$. So, the number of necklaces is the coefficient of $$a^3b^3$$ in the expansion of
$$\frac{1}{12}[(a+b)^6+3(a+b)^2(a^2+b^2)^2+4(a^2+b^2)^3+2(a^3+b^3)^2+(a^6+b^6)],$$
which is equal to $$\displaystyle \frac{1}{12}\left[\binom{6}{3}+3\binom{2}{1}\binom{2}{1}+2\binom{2}{1}\right]=3$$.
1. The cycle index polynomial of $$D_8$$ is $$\frac{1}{16}[x_1^8+4x_1^2x_2^3+5x_2^4+2x_4^2+4x_8]$$. So, the number of necklaces is the coefficient of $$a^4b^3c$$ in the expansion of
$$\frac{1}{16}\left[(a+b+c)^8+4(a+b+c)^2(a^2+b^2+c^2)^3+5(a^2+b^2+c^2)^4+2(a^4+b^4+c^4)^2+4(a^8+b^8+c^8)\right]$$
which is equal to $$\displaystyle \frac{1}{16}\left[\binom{8}{4}\binom{4}{3}+4\binom{2}{1}\binom{3}{2}\right]=19$$.