Let $f(x,y,z)=(x+y+z-1)^2-4xyz$. One can verify by inspection that for all $x,y,z$ $$f(x^2,y^2,z^2)=16 f\left(\frac{1+x}{2},\frac{1+y}{2},\frac{1+z}{2}\right)\cdot f\left(\frac{1-x}{2},\frac{1-y}{2},\frac{1-z}{2}\right).$$
I came to the above in the process of playing around with the formula $$\cos^2 t+\cos^2u+\cos^2v =1+2\cos t\cos u\cos v,$$ valid when $u+v+t=0$. (This in turn is a special case of the identity discussed in another recent question of mine.) This relation implies that $f(\cos^2 t,\cos^2 u,\cos^2v)=0$ when $u+v+t=0$. To connect this further with the desirerd relation, if we substitute $(x,y,z)=(\cos 2t,\cos 2u,\cos 2v)$, then the identity of interest may be shown to take the form $$f(\cos^2 2t,\cos^2 2u,\cos^2 2v)=16f(\cos^2 t,\cos^2 u,\cos^2 v)f(\sin^2 t,\sin^2 u,\sin^2 v).$$ From this we see that if $t+u+v=0$ iff $2t+2u+2v=0$ as well and therefore $f(\cos^2 t,\cos^2 u,\cos^2 v)=0$ iff $f(\cos^2 2t,\cos^2 2u,\cos^2 2v)=0$.
This verifies the desired identity in the case corresponding to $u+v+t=0$. What I want to know: Is there a geometric/trigonometric account of the general result?
