Given the condition [tex]\(a + b + c = 0\)[/tex], we are asked to prove the following identity:
[tex]\[
\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3
\][/tex]
### Step-by-Step Solution:
1. We start with the given condition:
[tex]\[
a + b + c = 0
\][/tex]
2. Consider the expression we want to prove:
[tex]\[
\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}
\][/tex]
3. Combine the fractions by multiplying by [tex]\(abc\)[/tex] (common denominator) to simplify:
[tex]\[
\left( \frac{a^2}{bc} \right) (abc) + \left( \frac{b^2}{ca} \right) (abc) + \left( \frac{c^2}{ab} \right) (abc)
\][/tex]
4. Simplify each term:
[tex]\[
\frac{a^2 abc}{bc} + \frac{b^2 abc}{ca} + \frac{c^2 abc}{ab}
\][/tex]
This reduces to:
[tex]\[
a^3 + b^3 + c^3
\][/tex]
5. Recall the identity for the sum of cubes and apply the given condition [tex]\(a + b + c = 0\)[/tex]:
[tex]\[
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
\][/tex]
6. Substitute [tex]\(a + b + c = 0\)[/tex] into the identity:
[tex]\[
a^3 + b^3 + c^3 - 3abc = 0
\][/tex]
Therefore, we have:
[tex]\[
a^3 + b^3 + c^3 = 3abc
\][/tex]
7. Substitute back into the simplified expression:
[tex]\[
\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = \frac{a^3 + b^3 + c^3}{abc}
\][/tex]
8. Since [tex]\(a^3 + b^3 + c^3 = 3abc\)[/tex], substitute this in:
[tex]\[
\frac{a^3 + b^3 + c^3}{abc} = \frac{3abc}{abc}
\][/tex]
9. Simplify the right-hand side:
[tex]\[
\frac{3abc}{abc} = 3
\][/tex]
Thus, we have proven that:
[tex]\[
\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3
\][/tex]
