Solving a system of equations $x^2 +y^2 −z(x+y)=2,y^2 +z^2 −x(y+z)=4,z^2 +x^2 −y(z+x)=8$

Question

I'm trying to solve the following system over $\Bbb R$:

$\begin{cases}{x^2 +y^2 −z(x+y)=2\\ y^2 +z^2 −x(y+z)=4\\ z^2 +x^2 −y(z+x)=8}\end{cases}$

Adding all the equations gives $2(x^2+y^2+z^2-xz-yz-xy)=14$. This doesn't look like $(x+y+z)^2$… Do you have some hints?

Answer 1

$$2(x^2+y^2+z^2-xz-yz-xy)=((x-y)^2 +(y-z)^2+(z-x)^2)$$ Subtracting first two equation give $$(z-x)(x+y+z)=2$$ Similarly develop other relation. $$\left(\frac{4}{(x+y+z)^2}+\frac{16}{(x+y+z)^2}+\frac{36}{(x+y+z)^2}\right)=14$$