Let $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$. Why does $a=b=c$? Let $a,b,c\in \mathbb{R}$  and $a,b,c \ne 0$ and  $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$.
Why does $a=b=c$?

$$2(a^2b^2+b^2c^2+c^2a^2)=2a^2bc+2b^2ca+2c^2ab\\\implies (a^2b^2+b^2c^2-2ab^2c)+(a^2b^2+c^2a^2-2a^2c)+(b^2c^2+c^2a^2-2abc^2)=0\\\implies (ab-bc)^2+(bc-ca)^2+(ca-ab)^2=0$$
From here you can only say $$ab=bc=ac$$ 
However since $a,b,c\neq0$ you get $a=b=c$
Bingo!
