<s>[INST] Nonlinear equations in 3 variables Solve for $x,y\ \mbox{and}\ z$:
$$
\left\{\begin{array}{rcr}  
x + y + z & = & 2  
\\[1mm]
\left(x + y\right)\left(y + z\right) + \left(y+z\right)\left(z+x\right) +
\left(z + x\right)\left(x + y\right) & = & 1  
\\[1mm]
x^{2}\left(y + z\right) + y^{2}\left(x + z\right) + z^{2}\left(x + y\right)
& = & -6
\end{array}\right.
$$
I tried this by adding the second and third equations, and then using the first equation to substitute values. However, this is getting nowhere.
 [/INST] Considering the equations$$x+y+z-2=0\tag 1$$ $$(x+y) (x+z)+(x+z) (y+z)+(x+y) (y+z)-1=0 \tag 2$$ $$x^2 (y+z)+y^2 (x+z)+z^2 (x+y)+6=0\tag 3$$ first eliminate $z$ from $(1)$ $$z=-x-y+2\tag 4$$ Replace in $(2)$ to get $$\left(-x^2+2 x+3\right)+(2-x) y-y^2=0\tag 5$$ Solve the quadratic in $y$ and keep the largest root; this gives $$y=\frac{1}{2} \left(\sqrt{-3 x^2+4 x+16}-x+2\right)\tag6$$ Plug $z$ and $y$ in $(3)$ (this is the tedious part) and get $$-3 x^3+6 x^2+9 x=0 \tag 7$$ So the possible roots $$x= -1\qquad ,\qquad x= 0\qquad ,\qquad x= 3$$
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