If $a, b, c$ are all non-zero and $a+b+c=0,$ prove that $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}=3$.

(N/A) Given that $a, b, c$ are non-zero and $a+b+c=0$.
We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
Since $a+b+c=0$,the right side becomes $0$,so $a^{3}+b^{3}+c^{3}-3abc = 0$,which implies $a^{3}+b^{3}+c^{3}=3abc$.
Now,consider the expression $\frac{a^{2}}{bc} + \frac{b^{2}}{ca} + \frac{c^{2}}{ab}$.
Taking the least common multiple $(LCM)$ of the denominators $bc, ca, ab$,which is $abc$,we get:
$\frac{a^{2}(a) + b^{2}(b) + c^{2}(c)}{abc} = \frac{a^{3}+b^{3}+c^{3}}{abc}$.
Substituting $a^{3}+b^{3}+c^{3} = 3abc$ into the expression:
$\frac{3abc}{abc} = 3$.
Thus,the expression equals $3$.