<s>[INST] Solve system:$yz(y+z-x)= a(x+y+z)\\ zx(z+x-y)= b(x+y+z)\\ xy(x+y-z)= c(x+y+z)$ Solve system:
$$yz(y+z-x)= a(x+y+z)\\
 zx(z+x-y)= b(x+y+z)\\
 xy(x+y-z)= c(x+y+z)$$
$$a, b, c>0$$
I can only solve the system at $a= 2, b= 5, c= 10$. I have tried all things but it's hard with me. Somebody help me?
 [/INST] Solve equation 1 for $x$:
$$ x=-{\frac { \left( y+z \right)  \left( -yz+a \right) }{yz+a}}$$
(unless $yz+a=0$, which I'll ignore for now).
Substitute this into the other two equations, and simplify.  I get
$$ \eqalign{yz \left( y+z \right)  \left( y{z}^{3}-ayz-a{z}^{2}-byz+{a}^{2}-ab
 \right)  &= 0\cr
yz \left( y+z \right)  \left( z{y}^{3}-a{y}^{2}-ayz-cyz+{a}^{2}-ac
 \right) &= 0\cr} $$
So either $y=0$, in which case $x = -z$, or $z=0$, in which case $x=-y$, or $y+z=0$, in which case $x=0$, or $$\eqalign{y{z}^{3}-ayz-a{z}^{2}-byz+{a}^{2}-ab &= 0\cr z{y}^{3}-a{y}^{2}-ayz-cyz+{a}^{2}-ac &= 0\cr}$$
Solve the first of these for $y$, substitute into the second, and simplify.
$$ \eqalign{y &= \frac{a (a-b-z^2)}{z(a+b-z^2)}\cr
\left( -{z}^{2}+a \right) & \left(c{z}^{6}+ \left( ab-2\,ac-2\,bc \right) {z}^{4}- \left( a+b \right) 
 \left( 2\,ab-ac-bc \right) {z}^{2}+ab(a-b)^2\right)
=0\cr}$$
Note the second factor in the left side of the last equation is a cubic in $z^2$.
Things may get somewhat messy if you solve this explicitly.  
</s>