Hypercube "sides"

Consider a n-cube , which has $2n\;$ (n-1)-faces , as for example a 2-cube , which is a square, has $4\;$ 1-faces , which is $4$ lines as sides. Or for example, a 3-cube , which is a regular cube, has $6\;$ 2-faces , which is $6$ squares as sides. This will be used to establish by induction what happens next. Given a n-cube , extend a copy of same into $n+1$ space. Then for each of the $2n\;$ (n-1)-faces of the n-cube , a new n-face is formed for the (n+1)-cube ....PLUS the original n-cube and its copy extended into $n+1$ space. That's a total of $2n+2\;$ n-faces for the new (n+1)-cube , and thus we see that this relation holds by induction.

For a hypercube, where $n=4$ , the number of (4-1=3)-faces is $2n=8$ .

A hypercube looks like this:

(Actually this is just a 2D projection of a hypercube, a four dimensional object)

Although hard to visualize, in four dimensions each of those "warped cubes" looks like a real cube, i.e. right angles, squares for sides, etc. The only two that don't look "warped" in 3D are the ones in the middle and outside. This is similar to how the faces of a cube look like proper squares in 3D, but become "warped" when projected onto two dimensions.

So, we can count the "cubes" in the above figure. One little one in the middle. One big one on the outside. And six "warped in 3D" ones that join each of the faces of the smaller one with the corresponding face of the bigger one.

$1+1+6 = \boxed8$