ABC26GN0988 · Simplification

$a+b+c=0 \Rightarrow a=-(b+c) \Rightarrow a^{2}=(b+c)^{2}$. $$\therefore \quad \begin{aligned} & \frac{a^{2}}{\left(a^{2}-b c\right)}+\frac{b^{2}}{\left(b^{2}-c a\right)}+\frac{c^{2}}{\left(c^{2}-a b\right)} \\ & =\frac{(b+c)^{2}}{(b+c)^{2}-b c}+\frac{b^{2}}{b^{2}+c(b+c)}+\frac{c^{2}}{c^{2}+b(b+c)} \\ & =\frac{(b+c)^{2}}{b^{2}+c^{2}+b c}+\frac{b^{2}}{b^{2}+c^{2}+b c}+\frac{c^{2}}{b^{2}+c^{2}+b c} \\ & =\frac{b^{2}+c^{2}+2 b c+b^{2}+c^{2}}{b^{2}+c^{2}+b c}=\frac{2\left(b^{2}+c^{2}+b c\right)}{b^{2}+c^{2}+b c}=2 \end{aligned}$$