If `a+b+c = 0` find `(a^(2))/(bc) + (b^(2))/(ca) + (c^(2))/(ab)=` ?

Answer

To solve the problem, we need to find the value of the expression \(\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}\) given that \(a + b + c = 0\).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ a + b + c = 0 \]

2. **Rearrange the equation**:
From the equation \(a + b + c = 0\), we can express one variable in terms of the others. For example, we can write:
\[ c = - (a + b) \]

3. **Substitute \(c\) into the expression**:
We will substitute \(c\) in the expression \(\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}\):
\[ \frac{a^2}{b(-a-b)} + \frac{b^2}{(-a-b)a} + \frac{(-a-b)^2}{ab} \]

4. **Simplify each term**:
- The first term becomes:
\[ \frac{a^2}{-b(a+b)} = -\frac{a^2}{b(a+b)} \]
- The second term becomes:
\[ \frac{b^2}{-a(a+b)} = -\frac{b^2}{a(a+b)} \]
- The third term becomes:
\[ \frac{(a+b)^2}{ab} = \frac{a^2 + 2ab + b^2}{ab} \]

5. **Combine all terms**:
Now, we can combine all the terms:
\[ -\frac{a^2}{b(a+b)} - \frac{b^2}{a(a+b)} + \frac{a^2 + 2ab + b^2}{ab} \]

6. **Find a common denominator**:
The common denominator for the first two fractions is \(ab(a+b)\):
\[ -\frac{a^3 + b^3}{ab(a+b)} + \frac{a^2 + 2ab + b^2}{ab} \]

7. **Combine the fractions**:
\[ \frac{-(a^3 + b^3) + (a^2 + 2ab + b^2)(a+b)}{ab(a+b)} \]

8. **Use the identity for \(a^3 + b^3\)**:
We know that \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\). Since \(a + b = -c\), we can substitute:
\[ -(a^3 + b^3) = c(a^2 - ab + b^2) \]

9. **Final simplification**:
After simplification, we find that:
\[ \frac{3abc}{abc} = 3 \]

### Conclusion:
Thus, the value of the expression \(\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}\) is:
\[ \boxed{3} \]