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
If you are allowed to rotate the assembled cube freely there is exactly one way to assemble it.

Proof. Let $\{0,1,2,3\}$ be the set of colors. Color each vertex of the assembled $2$-cube with the color of the $1$-cube it belongs to.  Vertices of the assembled cube having the same color are necessarily space-diagonally opposite. Assume that a vertex having color $0$ is facing you. Then  its three adjacent vertices have colors $1$, $2$, $3$, arranged either counterclockwise or clockwise. Now convince yourself of the following: If $1$, $2$, $3$ are arranged counterclockwise, then $1$, $2$, $3$ will be arranged clockwise if you look at the assembled cube with the opposite $0$-vertex facing you.$\qquad\square$

It is so far unclear how this problem can be meaningfully formulated for an $n\times n\times n$ assembled cube.
Yes think like this to make a cube max side is $2$ as $2^3=8$ qnd we have $8$ eight cubes. Now take tge base . The four cubes can be arranged in $4!$ ways in the base. Now as we want each colour on each face   $2$ cubes with colour different than base colour cubes can be arranged on top of two base cubes maybe row or column. So now they can be arranged in $2$ ways and same can be done with the cubes in other row or column .so these are $2$ ways. Hence total ways are $4!.2.2=96$ and may be number of colours in generalized formula should be $n^{1/3}$