If a, b and c are all non-zero

To prove,

$\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}=3$\

We know that, $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$=0\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \quad[\because a+b+c=0$, given $]$

$=0$

$\Rightarrow \quad a^{3}+b^{3}+c^{3}=3 a b c$

On dividing both sides by $a b c$, we get

$\frac{a^{3}}{a b c}+\frac{b^{3}}{a b c}+\frac{c^{3}}{a b c}=3$

$\Rightarrow$ $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}=3$ 

Hence proved.