Each vertex of a cube is to be labeled with an integer $1$ through $8$, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$
Note that the sum of the numbers on each face must be 18, because $\frac{1+2+\cdots+8}{2}=18$. 
So now consider the opposite edges (two edges which are parallel but not on same face of the cube);
they must have the same sum value too.
Now think about the points $1$ and $8$. If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$. However, this scenario would yield no solution for $[7,2]$, which is a contradiction. (Try drawing out the cube if it doesn't make sense to you.)
The points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9$.
Now we have $4$ parallel sides $1-8, 2-7, 3-6, 4-5$.
Thinking about $4$ endpoints, we realize they need to sum to $18$.
It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.
So if we fix one direction $1-8 ($or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$
Now, the problem is the same as arranging $4$ points in a two-dimensional square, which is $\frac{4!}{4}=\boxed{\textbf{(C) }6.}$