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 $a=b=c$?

Let $a=cx$ and $b=cy$.  The equation becomes $x^2y^2+y^2+x^2=xy(x+y+1)$, or
$$(y^2-y+1)x^2-(y^2+y)x+y^2=0$$
As a quadratic in $x$, the discriminant, which must be non-negative in order for $x$ to be real, is
$$(y^2+y)^2-4(y^2-y+1)y^2=-3y^2(y-1)^2$$
So if $y\not=0$, we must have $y=1$, which in turn implies $x=1$.
