<s>[INST] Solve for positive real solutions of cyclic equations $x+y^2+2xy=9$, $y+z^2+2yz=47$, $z+x^2+2xz=16$ 
Solve over positive reals $$x+y^2+2xy=9$$
$$y+z^2+2yz=47$$
$$z+x^2+2xz=16$$

With standard manipulation we get that $x+y+z=8$. Thus $x=8-y-z$ and we have two equations,
$$(8-y-z)^2+y^2+2(8-y-z)y=9$$
$$y+z^2+2yz=47$$
Which becomes
$$-y^2-2yz+15y-z=1$$
$$y+z^2+2yz=47$$
Adding up gets
$$-y^2+16y+z^2-z=1$$
$$\implies z^2=z+1+y^2-16y$$
And this substitutes to 
$$y+z+1+y^2-16y+2yz=47$$
$$\implies y+\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)+1+y^2-16y+2y\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)=47$$
However, this results into a nasty quartic which I don't know how to simplify. Thanks!
 [/INST] Continue by substituting $z=\frac{-y^2+15y-1}{2y+1}$ into $y+z^2+2yz=47$,
$$y+\left(\frac{-y^2+15y-1}{2y+1}\right)^2+2y\left(\frac{-y^2+15y-1}{2y+1}\right)=47$$
Simplify,
$$-3y^4+32y^3+69y^2-219y-46=0$$
and factorize
$$(y-2)(3y^3-26y^2-121y-23)=0$$
which yields the positive real solutions $x=1,\>y=2,\>z=5$. The other three sets of solutions are also real, but containing negative values.
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