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?

@iamvegan: thank you. For instance $(1)$-$(2)$ gives $(x-z)(x+z+y)=-2$, and I also get $(y-x)(y+x+z)=-4 ; (y-z)(y+z+x)=-6$. — Alphonse, Aug 22 '16 at 17:00
Computing the $S$-polynomials from the three polynomials gives $y^2-1=0,z+2y=0,x+y=0$. — Dietrich Burde, Aug 22 '16 at 19:11
@DietrichBurde: wow, this is a smart method! But could you explain how this works? I don't know what $S$-polynomials are. — Alphonse, Aug 22 '16 at 19:17
For some examples see here, section $4$. Of course, for algebra-precalculus this is not useful. — Dietrich Burde, Aug 22 '16 at 19:21

Aakash Kumar · Accepted Answer · 2016-08-23T07:12:58.847

2

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

edited Aug 23 '16 at 07:12

answered Aug 22 '16 at 17:01

Aakash Kumar

3,480

Ah yes, I didn't see the first one! Thank you. Then I can try $(z-x)/(y-x)=2/(-4)$ (provided that $x+y+z≠0$) and so on. I will try, thanks! – Alphonse Aug 22 '16 at 17:08
I loved the trick that gives $(x-y)^2 +(y-z)^2+(z-x)^2$ (+1) – iamvegan Aug 22 '16 at 17:33

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

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