Question:
If
\[ a+b+c=0, \] then \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} = \]
(a) 0
(b) 1
(c) -1
(d) 3
Solution:
Taking LCM:
\[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} = \frac{a^3+b^3+c^3}{abc} \]
Using identity:
\[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \]
Given:
\[ a+b+c=0 \]
Therefore,
\[ a^3+b^3+c^3-3abc=0 \]
\[ a^3+b^3+c^3=3abc \]
Substituting:
\[ \frac{a^3+b^3+c^3}{abc} = \frac{3abc}{abc} \]
\[ =3 \]
Hence, the correct answer is:
\[ \boxed{3} \]