Show that if $x + y + z = xy + xz + yz$ then $x ^ 2 * (1 - y ^ 2) + y ^ 2 * (1 - z ^ 2) + z ^ 2 * ( 1 - x ^ 2) = 2(x + y + z)(xyz - 1)$

Question

the question

Show that if $x + y + z = xy + xz + yz$ then $x ^ 2 * (1 - y ^ 2) + y ^ 2 * (1 - z ^ 2) + z ^ 2 * ( 1 - x ^ 2) = 2(x + y + z)(xyz - 1)$

the idea

First of all i though of breaking all the products and i got to:

$x^2-x^2y^2+y^2-y^2z^2+z^2-z^2x^2=2x^2yz+2xy^2z+2xyz^2-2x-2y-2z$

Then i squared the given equality and got to

$x^2+y^2+z^2+2(xy+yz+xz)=x^2y^2+y^2z^2+x^2z^2+2x^2yz+2xy^2z+2xyz^2$

I tried processing it, but got to nothing usegul. Hope one of you can help me! Thank tou!

Answer 1

$$x^{2}\cdot \left(1-y^{2}\right)+y^{2}\cdot \left(1-z^{2}\right)+z^{2}\cdot \left(1-x^{2}\right)-2 \left(x+y+z\right) \left(x y z-1\right)=-\left(x y+x z+y z+x+y+z+2\right)\cdot \underbrace{\left(x y+x z+y z-(x+y+z)\right)}_0=0$$