Dizzy algebra Part - II

$$ Let $$ $\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+a^2c^2}=k,$ $$ then $$ $\begin{aligned} a^4+b^4+c^4&=k(a^2b^2+b^2c^2+a^2c^2)\\ a^4+b^4+c^4-k(a^2b^2+b^2c^2+a^2c^2)&=0.&\;(1) \end{aligned}$ $$ By using Heron's formula , we obtain $$ $a^4+b^4+c^4-2(a^2b^2+b^2c^2+a^2c^2)=(a+b+c)(a-b-c)(a-b+c)(a+b-c).$ $$ Since $a+b+c=0$ , we have $$ $\begin{aligned} a^4+b^4+c^4-2(a^2b^2+b^2c^2+a^2c^2)&=0.&&&&&&&&&&\;(2) \end{aligned}$ $$ Comparing $(1)$ and $(2)$ yields $k=\boxed{2}$ .