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.
 A: Way:
$1)$ express LHS-s via elementary symmetric polynomials; 
$2)$ consider related cubic equation.

$1)$ Elementary symmetric polynomials for this problem are: 
$e_1(x,y,z) = x+y+z$;
$e_2(x,y,z) = xy+xz+yz$;
$e_3(x,y,z) = xyz$.
So,
$$x+y+z= e_1;$$
$$(x+y)(y+z) + (y+z)(z+x) + (z+x)(x+y)\\ =3(xy+yz+xz)+x^2+y^2+z^2 \\
=(x+y+z)^2+(xy+xz+yz) \\ = \left(e_1\right)^2 + e_2;$$
For the last equation: note that
$$(x+y+z)(xy+xz+yz) = x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+3xyz,$$
therefore
$$x^2(y+z) + y^2(x+z) + z^2(x+y) = e_1e_2-3e_3.$$
Now we have:
$$\left\{\begin{array}{l}e_1 = 2;\\(e_1)^2+e_2 =1; \\ e_1e_2-3e_3 = -6;\end{array} \right.
\Rightarrow
\left\{\begin{array}{l}e_1 = 2;\\e_2 =-3; \\ e_3 = 0.\end{array} \right.
$$
$2)$
So (according to Vieta's formulas), triple $(x,y,z)$ is solution of cubic  equation
$$w^3-e_1w^2+e_2w-e_3=0.$$
$$w^3-2w^2-3w=0.$$
$$w(w^2-2w-3)=0.$$
Hence, $(x,y,z)=(-1,0,3)$ and all its permutations.
A: 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 $$x= -1\qquad ,\qquad x= 0\qquad ,\qquad x= 3$$
