If \(a, b, c\) are all non-zero and \(a+b+c=0\), prove that \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3 \]

\[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} \] \[ = \frac{a^3+b^3+c^3}{abc} \] Since, \[ a+b+c=0 \] Using the identity \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] Therefore, \[ a^3+b^3+c^3-3abc=0 \] \[ a^3+b^3+c^3=3abc \] Hence, \[ \frac{a^3+b^3+c^3}{abc} = \frac{3abc}{abc} \] \[ =3 \] Therefore, \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3 \]