If `alpha,beta and gamma` are the roots of `x^3+qx + r = 0` then`sumalpha/(beta+gamma)` , is (i)`3` (ii)`q+r` (iii)`q/r` (iv) `-3`

Options

A. 3
B. q+r
C. `q//r`
D. `-3`

Answer

To solve the problem, we need to find the value of the expression \( \sum \frac{\alpha}{\beta + \gamma} \), where \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 + qx + r = 0 \).

### Step-by-Step Solution:

1. **Understanding the Roots**:
The roots of the polynomial \( x^3 + qx + r = 0 \) are \( \alpha, \beta, \gamma \). According to Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta + \gamma = 0 \) (since the coefficient of \( x^2 \) is 0).
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = q \).
- The product of the roots \( \alpha\beta\gamma = -r \).

2. **Rewriting the Expression**:
The expression we need to evaluate is:
\[ \sum \frac{\alpha}{\beta + \gamma} = \frac{\alpha}{\beta + \gamma} + \frac{\beta}{\alpha + \gamma} + \frac{\gamma}{\alpha + \beta} \]
Using the fact that \( \beta + \gamma = -\alpha \), \( \alpha + \gamma = -\beta \), and \( \alpha + \beta = -\gamma \), we can rewrite the expression as:
\[ \sum \frac{\alpha}{\beta + \gamma} = \frac{\alpha}{-\alpha} + \frac{\beta}{-\beta} + \frac{\gamma}{-\gamma} \]

3. **Simplifying Each Term**:
Each term simplifies to:
\[ \frac{\alpha}{-\alpha} = -1, \quad \frac{\beta}{-\beta} = -1, \quad \frac{\gamma}{-\gamma} = -1 \]
Therefore, we have:
\[ \sum \frac{\alpha}{\beta + \gamma} = -1 - 1 - 1 = -3 \]

4. **Conclusion**:
Thus, the value of \( \sum \frac{\alpha}{\beta + \gamma} \) is \( -3 \).

### Final Answer:
The correct option is (iv) \(-3\).