F(x,y,z) = x2+y2+z2-2xy-2yz-2zx, so
F(r2,s2,t2) = (r+s+t)(r+s-t)(r-s+t)(r-s-t).
P = [sin2(U/2),sin2(V/2),sin2(W/2)], with U+V+W = 2kπ.
We want to prove that
F(sin2(U/2),sin2(V/2),sin2(W/2)) = -4(sin(U/2)sin(V/2)sin(W/2))2
Since the arguments are squares, we may use the factorised from of F.
It is convenient to take the first two factors together to get
F1 = (sin(U/2)+sin(V/2))2 - sin2(W/2).
As U+V+W = 2kπ, sin(W/2) = ±sin((U+V)/2), but the term is squared.
Using standard reduction formulas,
F1 = 4 sin2(W/4){cos2((U-V)/4)-cos2((U+V)/4)}
As cos(2Q) = 2cos2(Q) - 1, we get
F1 = 2sin2(W/4){cos((U-V)/2)-cos((U+V)/2)}, which reduces to
F1 = 4sin2(W/4)sin(U/2)sin(V/2).
If we take the second two factors as F2, a similar calculation gives
F2 = -4cos2(W/4)sin(U/2)sin(V/2).
As sin(W/2) = 2sin(W/4)cos(W/4), the product of F1 and F2 is as stated.