Number of ways to color a cube with three white and three black sides Let the faces be labelled as: A, B, C, D, E, F; with bottom labelled as: A, and the top as: F. The vertical faces are labelled in the cck ordering, from the top, as: B, C, D,and E.
Note: Am using w to denote the white color, and b to denote the black color. Also, am considering the vertical faces as a different category than the horizontal faces. The vertical faces are those parallel to the $y$-axis, while the horizontal faces are parallel to the $x$-axis.
Then, the distinct ways are: 

*

*A-w, F-b, and two distinguishable ways to color the vertical faces as: 
a. B- w, C- w, D- b, E- b,
b. B- w, C- b, D- w, E- b.


*A-w, F-w, and any one of the four vertical faces as w.
Hence, a total of three ways given by: 1a, 1b, 2.
But, the answer states: two ways.
Similarly, for coloring with two white and four black sides, have ways:

*

*A-w, F-w, and all vertical faces of the black color.

*A-w, F-b, and any one of the vertical faces as white.

*A-b, F-b, and two ways to color the vertical faces as:
a. B- w, C- w, D- b, E- b,
b. B- w, C- b, D- w, E- b.

Hence, a total of four ways.
The book states the answer for this case as: $2\times 2=4.$
Here, though the answer matches; but how the split in the two cases of 4w-2b, 2b-4w, was done with each case having two choices, is unclear.
Seems, as if all six faces are taken as of the same type.
Then, with three white and three black faces, have distinct ways given by:

*

*A-w, B-w, C-w, D-b, E-b, F-b $=$ A-b, B-b, C-b, D-w, E-w, F-w

*A-w, B-w, C-b, D-w, E-b, F-b $=$ A-b, B-b, C-w, D-b, E-w, F-w,

*A-w, B-b, C-w, D-b, E-w, F-b$=$ A-b, B-w, C-b, D-w, E-b, F-w.

But, by the approach of considering all faces as of the same type; again get three ways.
 A: We have paints of $2$ colours: White (W) and Black (B). Using these $2$ paints we have to colour the cube on all sides. Let's do it:
$\bf 6$ sides with the same colour:
Here, we can paint the cube on all sides with either B or W.
Thus, $2$ ways to do so: $(6W)$ or $(6B)$.
Kindly check the note 1 at the end.
$\bf 5$ sides with the same colour:
Here, if we paint one side with 1 colour all the other sides must be painted with the other colour.
Thus, $2$ ways to do so: $(5W,1B)$ or $(5B,1W)$.

Above $2$ cases were easy as we needn't arrange the colours in itself.

$\bf 4$ sides with the same colour:
First let's look at the case of $(4B,2W):$
Here, we'll focus at how $2W$s can arrange themselves as that would address all the variations and is easier to handle too.
The $2W$s may either be on the opposite sides or not on the opposite sides. That's it! 
Kindly check note 2.
All we can do now is switch the colour which was getting painted on $2$ sides with the other colour and so we'll again get $2$ different cases.
Thus, all in all, we have $4$ cases: $(4W,2B-a)(4W,2B-o)(4B,2W-a)(4B,2W-o)$
('a' means adjacent and 'o' means opposite)
$\bf 3$ sides with the same colour:
Here, since both colours are painted on 3 sides so we needn't switch as now it wouldn't matter. 
While this case is easy enough to image, it's a bit harder to explain.
Imagine: Either all the sides with same colour are together (touching each other) or such that $2$ are on opposite sides and one connecting them (like a "U"-shape).
Explain: Here, paint the bottom with side, A with W, then the rest $2$ W's may come on opposite sides on the sides B and D or adjacent to each other like on sides A and B. But here, you need to be careful that painting on F and then on one among B,C,D,E is not a different case. (In the same way as after painting on A and then painting on B,C or C,D or D,E or E,A aren't same from what we've already considered.)
(Here, your 1b and 2 are the same.)
Thus, the total number of ways: $(3W,3B-a)(3W,3B-o)$
$\bf 2/1/0$ sides with the same colour:
These are apparently repetitions of the cases and thus redundant.
Final answer:
Grand total $=2+2+4+2\bf=10$.

Note 1:
Yes, the colours for the given question can be switched! This is apparent from the $2$ cases that we are getting for $6$ sides with same colour. Which wouldn't have been possible if the majority colour was required to always be of one type. 
And you are trying to somehow differentiate them on the basis of the axes but that's not the case and generally speaking too, with a cube, that would almost never be the case. And if axes are to be made relevant then question would instead give you a cuboid with all $3$ dimensions of different lengths.
Note 2:
Here, your (2 and 3a) and (1 and 3b) are the same. How?
Notice that once whether AF or BD get painted with white, they'll become indistinguishable. You yourself can't, later after a day, tell whether you painted on AF side or on BD. From my comment, if the faces themselves were different by virtue of their (say) material so that you can touch and differentiate, then that would have been different. But we aren't given such a thing.
A: The referenced question allows arbitrary rotations of the cube. In particular, we do math and not physics and thus have no way to use gravity to distinguish the vertical direction from other directions. :)
Clearly, there is only one way to colour all faces black. The solution writes $2\cdot1$ because they count „all black“ together with „all white“ under this colour type.
There is only one way to have five black and one white face: You can rotate the cube to bring the white face to the top, for example. They write $2\cdot1$ again to account for five white and one black.
Two white faces can be adjacent or not. In the adjacent case you can rotate one of the white faces to the top and then by rotating around the vertical axis bring the other white face to the front, so the adjacent case is only one single configuration. The non-adjacent case is also only one configuration: if we rotate one white face to the top, the other is automatically at the bottom. Thus the answer stares $2\cdots2$ with one factor coming from swapping black and white.
For three white and three black faces, the „colour swap factor“ is $1$ because three white and three black is the same as three black and three white. If among the white faces, two are non-adjacent, then we can be rotated to become top and bottom face (cf. above). Then a $ z$-rotation can bring the third face to the front. If no two white faces are non-adjacent, rotate any white face to the top. Then the bottom is black. Hence you can bring one of the other white faces to the front by a $z$-rotation. Then the back face is black and the third face must be left or right. These two situations can be identified via a 90 degree $z$-rotation. So there is only one way to have the three white faces without an opposing pair. Therefore, the solution states $1\cdot2$.
