Puzzle About Cubes (from the book thinking mathematically) I want to confirm my solution to the given problem (solutions were not available in the book)

I have eight cubes. Two of them are painted red, two white, two blue
  and two yellow, but otherwise they are indistinguishable. I wish to
  assemble them into one large cube with each color appearing on each
  face. In how many different ways can I assemble the cube?



*

*The answer i got was 96 is this correct? 

*I also tried to generalize
    the question such that given $n$ cubes and $\sqrt n$ number of colors, I came up with 



$$n!\cdot2(n-\sqrt{n})!\cdot(n-2\sqrt{n})!\cdot2(n-(4\sqrt{n}-4))!$$

is the above generalization correct?
Thanks
EDIT
well to further explain my question, the rational behind the answer to the 1st question was we have 6 faces and if we take one of the faces we have 4! possibilities to arrange the 4 colors. that gives us 24 possibilities. and if move to the other faces, we have 2 faces of 2! possibilities and 3 faces of 1! possibilities thus total arrangements are 4!x2!x2!x1!x1!x1! = 96
for the second question yes the colors should be n^1/3 but for the generalization n should be  no of colors.
for example if we take the current question 
n=4


*

*if we take the front face as the base face we have n! = 24

*if we take the left face because of step 1 we have $(n-\sqrt n)! =2$

*if we take the back face because of step 2 we again have $(n-\sqrt
    n)! =2$

*if we take the right face because of step 1 and step 3 we have
$(n-2\sqrt n)! =1$

*if we take the top face because of step 1,2,3,4  we have $(n-(4\sqrt
    n -4))! =1$

*if we take the bottom face again because of step 1,2,3,4  we have
$(n-(4\sqrt n -4))! =1$


thus the result 4!x2!x2!x1!x1!x1! = 96

I will try to give a simple proof, by using the techniques taught in the book itself.
ENTRY Phase

*

*Draw a diagram



*

*Write down what you know

*

*There are 8 cubes which are identical to each other, except for their colors. There are 2 red, 2 green, 2 blue and 2 orange cubes.



*Be clear on what you want

*

*We want to assemble as many larger 2X2 cubes from these 8 smaller cubes with the following constraint - each color should appear on each face.

*Goal is to find the number of assemblies possible.



Specialize
Can you solve one particular case? Yes - as shown in the picture.
What else you notice? You could observe that cubes with same color are at diagonally opposite corners of the cube.
Also observe a hidden constraint that for any face to have each of the 4 colors, you need to have each of the color on all of the faces.
Attack
Can you conjecture based on the observation?
Specialize further by refining.
Conjecture:
To satisfy the given constraint, cubes of similar color has to be on diagonally opposite corners of the cube.
Visual Proof:
Let us represent positions of the smaller cubes using following matrix, with first matrix for the front face and second one for the rear face, both read from left to right.
$$\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix} \begin{bmatrix}5 & 6\\7 & 8\end{bmatrix}$$

*

*If 2 cubes are next to each other on the same row in a face, that will violate the constraint on two faces. So (1,2) or (3,4) cannot be of same color. By symmetry, this is true for rear face as well. ie, (5,6) or (7,8) also can't be of same color.

*If 2 cubes are diagonally opposite to each other on the same face, that will violate the constraint on that face. So (1,4) or (2,3) cannot be of same color. By symmetry, this is true for rear face as well. ie, (5,8) or (6,7) also can't be of same color.

*By symmetry after rotation, it can be argued that 1,6 can't be of same color. Similarly, (2,5), (3,8) and (4,7) can't be of the same color.

*Therefore same color pieces have to be in positions (1,8), (2,7), (3,6) and (4,5).

If we fix a given color, say blue at position 1, there are 24 combinations possible. If we swap blue with 3 other colors, there are another $24 \cdot 3$ possibilities.
So a total of $24 \cdot 4 = 96$ combinations are possible, which is the final answer.
