We can factor these out like this: $$(x+y)(x-y)=2(z+1)(x+y)$$and repeat it for the other equations. There are two cases, one where $x+y,y+z$, or $z+x$ equal 0, and one where none of them equal 0. Case 1: none of them are 0. We can divide to get $$x-y=2(z+1)$$$$y-z=2(x+1)$$$$z-x=2(y+1)$$You can solve this system of equations to get $x+y+z=-3$ and after solving the equations the only solution for this case is $(x,y,z)=(-1,-1,-1)$. Case 2: at least one of x+y y+z z+x is 0 so you cannot divide. These equations are basically the same so let's pretend x+y=0, or x=-y. Then the last two equations are $$x^2-z^2=2(-x^2+zx-x+z)=2(z-x)(x+1)$$$$-x^2+z^2=2(-zx-x^2+z+x)=2(z+x)(-x+1)$$ For the second equation we cannot divide if z+x is 0. Splitting into 2 more cases: Case 2a: x+z is also 0, so x=-z. Looking at the middle equation $$0=2(-2x)(x+1)$$Now we see x=0 or -1, so our solutions are $(x,y,z)=(0,0,0)$ or $(x,y,z)=(-1,1,1)$. Case 2b: x+z is not 0. We add the two equations to get: $$0=-4x^2+4z$$This means $x^2=z$. Substituting we get $$z-z^2=2(-z+zx-x+z)=2x(z-1)$$$$z^2-z=2(-zx-z+z+x)=2x(1-z)$$These are both the same equation, which is: $$z(z-1)=-2x(z-1)$$Then, either $z=1$ or $-2x=z$. This gives $(x,y,z)=(1,-1,1)$ or $(x,y,z)=(-2,2,4)$ when combined with the equation $x^2=z$. Overall, we have these solutions because the equations are cyclic: $$(x,y,z)=(-1,-1,-1)$$$$(x,y,z)=(0,0,0)$$$$(x,y,z)=(-1,1,1),(1,-1,1),(1,1,-1)$$$$(x,y,z)=(-2,2,4),(2,4,-2),(4,-2,2)$$