ABC26GN1002 · Simplification

$$\begin{aligned} & a b+b c+c a=0 \Rightarrow a b=-b c-c a, b c=-a b-c a, \\ & c a=-a b-b c . \\ & \begin{aligned} \therefore \quad \frac{1}{a^{2}-b c} & +\frac{1}{b^{2}-c a}+\frac{1}{c^{2}-a b} \\ & =\frac{1}{a^{2}+a b+a c}+\frac{1}{b^{2}+a b+b c}+\frac{1}{c^{2}+b c+c a} \\ & =\frac{1}{a(a+b+c)}+\frac{1}{b(a+b+c)}+\frac{1}{c(a+b+c)} \\ & =\frac{b c+c a+a b}{a b c(a+b+c)}=0 .[\because a b+b c+c a=0] \end{aligned} \end{aligned}$$