<s>[INST] Solve the system of equations $x + y^2 + z^3 = 3; y + z^2 + x^3 = 3; z + x^2 + y^3 = 3$ (where x,y,z are positive real numbers) It seems that $x=y=z=1$ is the only solution, but I can't find how to prove it. I've tried defining $a=x-1; b=y-1; c=z-1$, but with no success. Any ideas?
Geometrically, each equation defines a 3D surface. All 3 surfaces are identical, except they are oriented along different axes. We're looking for their intersection.
 [/INST] The system can be written as:
$$
\begin{align}
\begin{cases}
\;\; (x-1) &+\;\; (y-1)\,(y+1) &+\;\; (z-1)\,(z^2+z+1) &= 0
\\ \;\; (x-1)\,(x^2+x+1) &+\;\;(y-1) &+\;\; (z-1)\,(z+1) &= 0
\\ \;\; (x-1)\,(x+1) &+\;\;(y-1)\,(y^2+y+1) &+\;\; (z-1) &= 0
\end{cases}
\end{align}
$$
Regarding it as a linear homogeneous system in $x-1, y-1, z-1$, its determinant is:
$$
\Delta = 
\begin{vmatrix}
   1 &y+1 &z^2+z+1
\\ x^2+x+1 &1 &z+1
\\ x+1 &y^2+y+1 &1
\end{vmatrix}
$$
After routine calculations courtesy WA, and using that $x,y,z \gt 0\,$:
$$
\begin{align}
\Delta &= x^2 y^2 z^2 + x^2 y^2 z + x^2 y^2 + x^2 y z^2 + x^2 y z + x^2 z^2 + x^2 z + x y^2 z^2
\\ &\;\;\;\; + x y^2 z + x y^2 + x y z^2 + 2 x y z + x y + x z + y^2 z^2 + y z^2 + y z
\\ &\gt 0
\end{align}
$$
It follows that the only solution is $x-1=y-1=z-1=0\,$.
</s>