If a+b+c=0, the value of `((a^(2))/(bc)+(b^(2))/(ca)+(c^(2))/(ab))` is

Options

A. 2
B. 3
C. 4
D. 5

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. **Substitution**:
Since \(a + b + c = 0\), we can express one variable in terms of the others. For example, we can write \(c = -a - b\).

2. **Substituting \(c\)**:
Substitute \(c\) in the expression:
\[ \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = \frac{a^2}{b(-a-b)} + \frac{b^2}{(-a-b)a} + \frac{(-a-b)^2}{ab} \]

3. **Simplifying Each Term**:
- The first term becomes:
\[ \frac{a^2}{b(-a-b)} = -\frac{a^2}{b(a+b)} \]
- The second term becomes:
\[ \frac{b^2}{(-a-b)a} = -\frac{b^2}{a(a+b)} \]
- The third term becomes:
\[ \frac{(-a-b)^2}{ab} = \frac{(a+b)^2}{ab} \]

4. **Combining the Terms**:
Now, we can combine these terms:
\[ -\frac{a^2}{b(a+b)} - \frac{b^2}{a(a+b)} + \frac{(a+b)^2}{ab} \]

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

5. **Using the Identity**:
Recall that \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\). Since \(a + b = -c\), we can express this in terms of \(c\):
\[ a^3 + b^3 = -c(a^2 - ab + b^2) \]

6. **Final Expression**:
Now, substituting back into our expression, we can simplify and find that:
\[ \frac{(a+b)^2 - (a^3 + b^3)}{ab(a+b)} = \frac{c^2 + 3abc}{ab(-c)} \]

7. **Result**:
After simplification, we find that:
\[ \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3 \]

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