A question on colouring cubes We are given 6 distinct colours and a cube.We have to colour each face with one of the  six colours and two faces with a common edge must be  coloured with different colours.How many distinct colouring ways are there?
 A: Just for fun, here's a solution without Burnside's lemma.
A color can be used at most twice (on two opposite faces), so the number of colors used is $3,4,5$, or $6$. We count the number of colorings for each number of colors.
$3$ colors: $\binom63=20$ ways to choose the colors, it's easy to see that all ways of applying them to the cube are rotationally indistinguishable, so $20$ ways.
$4$ colors: $\binom62\binom42=90$ ways to choose the colors (two to be used twice and two to be used once), all colorings with the chosen colors are equivalent, so $90$ ways.
$5$ colors: $\binom61\binom54=30$ ways to choose the colors. Paint two opposite faces "red" (the color chosen to be used twice) and place the cube on a table with a red side on top and bottom; the other $4$ colors are to be applied to the sides. There are $3!$ circular permutations (since the cube can be turned on its vertical axis), but we divide that by $2$ since the cube can be turned upside down, so $30\cdot3=$ 90 ways.
$6$ colors: Paint one face red. Choose a color for the opposite face ($5$ choices). Paint the other faces with the other $4$ colors (circular permutations, $3!$ choices), so $5\cdot6=$ $30$ ways.
The final snswer is $20+90+90+30=230$.
With a palette of $N$ colors, the number of indistinguishable colorings is
$$\binom N3+\binom N2\binom{N-2}2+3\binom N1\binom{N-1}4+30\binom N6$$
$$=\frac1{24}\left[N(N-1)(N-2)(N^3-9N^2+32N-38)\right]$$
which agrees with Marko Riedel's solution using Burnside's lemma.
A: Caveat:  what  follows  could  not  be verified  at  the  OEIS  or
elsewhere for that matter.   Verification by the reader, computational
e.g. by a script or otherwise is invited. 
Remark. No longer necessary, since a verification without Burnside
has been provided.

Suppose  we treat  the problem  of coloring  a cube  with at  most $N$
colors  where  adjacent  faces may  not  have  the  same color  up  to
isomorphism under rotations. We intend  to use Burnside. To do this we
need  to iterate  over all  permutations  in the  group of  rotational
automorphisms of the  cube and compute the number  of proper colorings
that are fixed by each permutation.

There are $24$  permutations in this group. First,  rotations about an
axis passing through opposite vertices. No proper coloring is fixed by
these rotations because  the faces on the two  three-cycles would have
to have  the same color but  they are adjacent, for  a contribution of
zero. Second, rotations about an axis passing through the midpoints of
opposite edges.  These  exchange the two faces incident  on one of the
two edges which would have to have the same color in order to be fixed
under  these rotations  but this  cannot occur  in a  proper coloring,
again  for a  contribution of  zero.  Third, rotations  about an  axis
passing through the midpoints of opposite faces. The 90 degree and 270
degree rotations  create four-cycles  of adjacent faces,  which cannot
be the same color in a proper coloring, again giving zero.

This leaves just two types of permutations, the identity and rotations
about an axis  passing through the midpoints of  opposite faces by 180
degrees,  which exchange  opposite faces  on  the ring  of four  faces
between the two faces connected  by the axis. Colorings that are fixed
by this  rotation must have the same  color on the two  pairs of faces
being exchanged and as these pairs are adjacent the two colors must be
different. This gives $N\times (N-1)$ choices. The two faces connected
by the axis can be colored independently with any one of the remaining
$N-2$ colors,  giving $N(N-1)(N-2)^2,$ but there are  three such axes,
for a total of
$$3\times N(N-1)(N-2)^2.$$

The identity is the only  one remaining. The number of colorings fixed
by this  permutation is simply the  number of proper vertex colorings of the
octahedron, which is given by the chromatic polynomial see e.g. 
MathWorld
$$N(N-1)(N-2)(N^3-9N^2+29N-32).$$

It now follows  by Burnside that the number  of proper colorings under
rotation is
$$\frac{1}{24}
\left(3\times N(N-1)(N-2)^2 + N(N-1)(N-2)(N^3-9N^2+29N-32)\right).$$

This gives the sequence
$$0, 0, 1, 10, 55, 230, 770, 2156, 5250, 11460, 22935, 42790, 75361,
\ldots$$


There are many more related links at 
MSE Meta on Burnside/Polya.
A: You could consider this is a direct application of http://en.wikipedia.org/wiki/Burnside%27s_lemma
What you will do is figure out your group of permutations generated by your possible rotations.  $\mathscr{G} = \{e, R, RR, RRR, F, FF, FFF, FR, RF, \cdots\}$ where $R$ represents rotation to the right and $F$ represents rotation to the front (This may take a bit of time to figure out the entire group since it is not abelian; $RFRF \neq RRFF$ for example).  Consider how many possibilities are left unchanged by each element of the group.  Denote the colorings which are not changed by permutation $\phi$ as $X^\phi$.
$|X/\mathscr{G}| = \frac{1}{|\mathscr{G}|}\sum_{\phi\in\mathscr{G}}|X^\phi$|
Using this you can answer the even harder question of if adjacent faces are allowed to be the same color.
Notice however that in this specific case, since you have that adjacent faces must be different colors, the only time two colors can be the same are if they are opposite sides, so that makes our work much easier and we know that for all $90^\circ$ or $270^\circ$ turns there will be no possibilities that keep the coloring the same.  You should check then the cases of $e, RR, FF,$ and $RRFF$(which is the same as FFRR) (and any others which I have missed, though I think those are the only four cases).
In the case of $RR$, you could have:
$~~~~~~c_4$
$c_1 ~~ c_2 ~~ c_3$
$~~~~~~c_5$
$~~~~~~c_6$
where $c_1$ must be the same as $c_3$, and $c_2$ must be the same as $c_6$.
Choose color of $c_1$ (6 choices), choose color of $c_2$ (5 remaining choices), choose color of $c_5$ (4 remaining choices), choose color of $c_4$ (4 remaining choices, the choice of $c_5$'s color does not affect $c_4$'s available choices).  In doing so, we force $c_3$ to be the same as $c_1$ and we force $c_6$ to be the same as $c_2$.  Thus, there are $6\cdot 5\cdot 4\cdot 4$ possible colorings which are preserved under $RR$ and $|X^{RR}| = 6\cdot 5\cdot 4\cdot 4$
Do similarly for the other cases.
A: There are only 30 distinguishable colourings. Fix one face with the first colour eg black and choose the opposite face colour in 5 ways. The remaining four faces form a ring. When you fix one colour in this ring, the other colours can be arranged in $3!$ ways.
