Can anyone show me, how to solve these system of Equations: $$\begin{align*}
x+y+z &= 2 \\
(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y) &= 1 \\
x^2(y+z)+y^2(z+x)+z^2(x+y) &= -6
\end{align*}$$
Can anyone explain me the solution. I asked it in mathoverflow but god knows why it was closed. Please help and detail the process in step by step. It is a RMO question (The Pre- Indian team selection exam for IMO)
 A: So I'll assume you're only looking for integer solutions. Call the equations (1), (2), and (3) respectively. Expand (2):
$ x^2 + y^2 + z^2 + 3x y + 3 x z + 3 y z = 1.$
From this equation, subtract the square of (1) to obtain
$ x y + x z + y z = -3.$
Subtract double this expression from the square of (1) to obtain
$ x^2 + y^2 + z^2 = 10.$
Aha! Since $x,y,z$ are integers, this means $x=\pm 3, y=\pm 1, z =0$ or some permutation thereof. (Just look at the various cases. There aren't that many.) By (1), this means $x=3,y=-1,z=0$ or some permutation thereof.
A: $$x+y+z=2$$
$$x^2+y^2+z^2+2(xy+zx+yz)=4$$
The second equation is $x^2+y^2+z^2+3(xy+zx+zy)=1 \implies zx+yz+yx=-3$
You get $x^2+y^2+z^2=10$
Using the $x+y+z=2$ in the third equation.
$x^2(2-x)+y^2(2-y)+z^2(2-z)=-6 \implies 2(x^2+y^2+z^2)-x^3+z^3+z^3=-6$
A little re-arranging would give 
$26 =x^3+y^3+z^3 $
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$26-3xyz=(2)(10-(-3)) \implies 26-26=3xyz \implies $ one of the $x,y,z$ is zero.
Consider $z=0$ (Generality is not lost) Now you have $x^2+y^2=10$ and $yx=-3$
You get solutions as $(x,y,z)=(3,-1,0),(0,-1,3)(0,3,-1),(-1,0,3),(3,0,-1)$ and $(-1,0,3)$.
A: Generally proceed as:


*

*Is calculated $s=x+y+z, q=xy+yz+zx, p=xyz;$

*Solve the equation $t^3-st^2+qt-p=0.$

