Solve for real numbers. $$ \begin{cases} 6x+3y+2z=18 \\ 108(x+y+z)^{(x+y+z)}=6^{(x+y+z)}xy^2z^3 \end{cases} $$
To solve this system of equation by trial and error I found $(x,y,z) = (1,2,3)$. I could not find any other root.
From second equation we see $xz>0$ , we also know $x+y+z>0$. I tried to show there are no other roots, please help me to complete it.