<s>[INST] How to solve a system of three nonlinear equation in a simple way Given the system:
$$
  \begin{cases}
   x^2y^2+x^2z^2=axyz &  \\
   y^2z^2+y^2x^2=bxyz &\\
   z^2x^2+z^2y^2=cxyz      
  \end{cases}
$$
The solution could be gotten in a very tedious way. Is it possible to solve it considering some symmetry property of the system if any? Thanks in advance. 
 [/INST] If is clear we cannot have some  $a,b,c$ positive while the rest negative. Otherwise,
$xyz = 0$ and we only have trivial solution.
For simplicity, I will assume $a, b, c > 0$. 
If any one of $x,y,z$ vanishes, say $x = 0$, it is clear the equations implies $y z = 0$. This means if one of $x,y,z$ vanish, at least two of $x,y,z$ vanish. If two of $x,y,z$ vanish, then the third one is arbitrary.
To eliminate this sort of trivial solutions, we will first assume $x, y, z > 0$.
If we divide everything by $x^2 y^2 z^2$, we have
$$\begin{cases}
\frac{1}{z^2} + \frac{1}{y^2} &= \frac{a}{xyz}\\
\frac{1}{x^2} + \frac{1}{z^2} &= \frac{b}{xyz}\\
\frac{1}{y^2} + \frac{1}{x^2} &= \frac{c}{xyz}
\end{cases}
\quad\iff\quad
\begin{cases}
\frac{1}{x^2} &= \frac{b+c-a}{2xyz}\\
\frac{1}{y^2} &= \frac{a+c-b}{2xyz}\\
\frac{1}{z^2} &= \frac{a+b-c}{2xyz}
\end{cases}
\tag{*1}
$$
This implies
$$\begin{align}
\frac{1}{(xyz)^2} &= \left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)\frac{1}{(xyz)^3}\\
\iff\quad\quad xyz &= \left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)
\end{align}
$$
Substitute this back into $(*1)$ immediately give us
$$\begin{cases}
x &= x_0 \stackrel{def}{=} \sqrt{\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)}\\
y &= y_0 \stackrel{def}{=} \sqrt{\left(\frac{b+c-a}{2}\right)\left(\frac{a+b-c}{2}\right)}\\
z &= z_0 \stackrel{def}{=} \sqrt{\left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)}
\end{cases}\tag{*2}$$
What happens if some or all of the $x,y,z$ are negative? If one look back the original
set of equations, the LHS is positive. Under the assume that $a, b, c > 0$, we find $xyz$ is positive. This means in general, the set of non-trivial solutions of the original
equation is given by that in $(*2)$ with an even number of $x,y,z$ flipped sign. i.e. 
$$(x,y,z) = (x_0,y_0,z_0), (x_0,-y_0,-z_0), (-x_0,y_0,-z_0) \text{ or } (-x_0,-y_0,z_0)$$
</s>