Tricky simultaneous Equations

Question

Given the simultaneous equations $$ x + yz = y + zx = z + xy $$

and

$$x^2 + y^2 + z^2 = 6$$

show that $x = 1$ or $y = z$ and hence solve the equations.

Answer 1

Let $$z+yz=t,y+zx=t,z+xy=t$$ and we get (from the last equation) $$z=t-xy$$, using this in (1) and (2) $$x+y(t-xy)=t$$ $$y+x(t-xy)=t$$ solving the last equation for $y$: $$y(1-x^2)=t(1-x)$$ Can you proceed? Factorizing: $$(1-x)(y(1+x)-t)=0$$ Substituting $x=1$ into the system: $$1+yz=y+z$$ or $$(1-z)(1-y)=0$$