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

Let's see.  Label the eight positions FLB, FLT, FRB, FRT, bLB, bLT, bRB, bRT.
There 4 possible colors for FLB, 3 for FLT, 2 for FRB, 1 for FRT.
bLB cannot share its color with any L, of **B so the only color it can have is that of FRT.  Each of the remaining three slots can only have the color of its opposite corner (Otherwise it will share a color with one of its three faces).
So there are 4! to color a 2x2 cube.  Now there are 2 of each color so there are $2^4$ ways to chose which of the two cubes for each slot.  
So there are $4!*2^4 = 24*16 = 384$.
But this distinguishes between rotations.  If I distinguish that I will always start with, arbitrarily the first yellow cube in FLB position, and it must be adjacent to all three of the other colors and if I distinguish that, arbitrarily, a blue cube must be in FRB, then this is reduced by 8 (where the first yellow cube goes is 8 and what color to choose for FRB is 3) then there are 16 possibility.
If further the 2 cubes of the same color are indistinguishable then I must reduce by the choices of which blue, white and red cubes.  Or $2^3$ leaving only 2 possibilities.  
If we allow symmetries then there is only one possibility.
But if that were the interpretation there'd be simpler way to do the puzzle.  The Yellow must go somewhere.  On one of the faces the the yellow will be next to the blue (and on another face next to the yellow and the third next to the white).  Orientate ourselves to that face and rotate it so the blue is to the right of the yellow.  Our only choice is whether the red or the white should go above the yellow.  This is essentially choosing whether we want a "right or left handed orientation".
To extend to an $n \times n \times n$ cube is different problem entirely as there are center unviewed cubes and not all cubes are on all three face.  
To extend a 2 cube to the n-th dimension would be $2^{n-1}!*2^{2^{n-1}}$ and with respect to orientation $2^{n-1}!*2^{2^{n-1}}/2^n*n!$.  (I think).
If we don't distinguish between individual cubes of the same color, I think, this will be $2^{n-2}$ which I believe are the number of left-handed/right-handed orientations.  Or maybe it is only 2.  
