Algebraic proof to show product of three numbers is zero from a system Let $a, b, c$ be real numbers satisfying: $$(a+b)(b+c)(c+a)=abc$$
$$(a^3+b^3)(b^3+c^3)(a^3+c^3)=a^3b^3c^3$$
Prove that $abc=0$
My work: I tried to prove it by contradiction, assuming $abc\neq0$ so $a,b,c\neq0.$
Then I factored second equation to be $(a+b)(b+c)(a+c)(a^2-ab+b^2)(b^2-bc+c^2)(a^2-ac+c^2)=(abc)^3$. Then when you divide the second equation by the first equation, you get $(a^2-ab+b^2)(b^2-bc+c^2)(a^2-ac+c^2)=a^2b^2c^2$. At this point I think there maybe a AM-GM inequality I can use, but I'm stuck. Any suggestions or alternate ways to prove this?

Look at the first equation as a quadratic equation in $a$.  It has discriminant $(b^2+bc+c^2)^2-4bc(b+c)^2=(b^2-bc+c^2)-8b^2c^2$.  So we have
$$
\begin{align*}
a^4b^4c^4&=(a^2b^2c^2)^2\\
&=\prod_{cyc}(b^2-bc+c^2)^2\\
&\geq\prod_{cyc}8b^2c^2\\
&=512a^4b^4c^4
\end{align*}
$$
which only works if $abc=0$.
