Q:

"Point P lies on the sphere described a cube."

Point P lies on the sphere described a cube. Show that the sum of
  squared distances of the point P of the vertices of the cube does not
  depend on the choice of P.

I cannot found any logical solution of that - I know that this is true, but I don't know why. 

A:

We assume that the sphere and cube are centered at the origin, and the cube has the 8 coordinates $(\pm 1, \pm 1, \pm 1)$.
Let $A,B,C$ denote the unit vectors along each axis, so that the vertices of the cube are $\pm A \pm B \pm C$.  Now we must compute the sum, over all 8 possible choices of $\pm$, of $(P\pm A\pm B\pm C)^2$, where multiplication denotes the inner product.
It's not so hard to do this, since most of the terms disappear due to symmetry.  In particular, you can see right away that the linear term in $P$ is just $P\sum (\pm A \pm B \pm C) = 0$.
We can even finish the calculation if we want.  Everything cancels except $8(P^2 + A^2 + B^2 + C^2) = 32$.  We can even double-check our answer by letting $P=A$, for example, and computing $2+2+2+2+6+6+6+6=32$.
Note that the final quantity only depends on the distance from $P$ to the origin.  So this actually works for a sphere of any radius, regardless of the size of the cube!  For example, if $P$ is a vertex of the cube, the formula should give us $4+4+4+8+8+8+12=48$ (direct computation).  And indeed, $8(3+1+1+1)=48$ (plugging in to $8(P^2 + A^2 + B^2 + C^2)$).

