If ` alpha , beta , gamma ` are the roots of `x^3 +qx +r=0` then ` sum ( beta + gamma )^(-1)`=

Options

A. `(q)/(r )`
B. `(r )/(q)`
C. `-(q)/( r)`
D. `-r/q`

Answer

To solve the problem, we need to find the value of the summation of the inverses of the sums of the roots of the polynomial equation \(x^3 + qx + r = 0\), where the roots are \(\alpha\), \(\beta\), and \(\gamma\).

### Step-by-step Solution:

1. **Identify the Roots**:
The roots of the polynomial \(x^3 + qx + r = 0\) are given as \(\alpha\), \(\beta\), and \(\gamma\).

2. **Use Vieta's Formulas**:
From Vieta's formulas for a cubic equation \(ax^3 + bx^2 + cx + d = 0\):
- The sum of the roots \(\alpha + \beta + \gamma = -\frac{b}{a}\).
- The sum of the products of the roots taken two at a time \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\).
- The product of the roots \(\alpha\beta\gamma = -\frac{d}{a}\).

For our equation:
- \(a = 1\), \(b = 0\), \(c = q\), \(d = r\).
- Therefore, \(\alpha + \beta + \gamma = 0\),
- \(\alpha\beta + \beta\gamma + \gamma\alpha = q\),
- \(\alpha\beta\gamma = -r\).

3. **Express \(\beta + \gamma\)**:
Since \(\alpha + \beta + \gamma = 0\), we can express \(\beta + \gamma\) as:
\[ \beta + \gamma = -\alpha \]
Similarly, we can express the other pairs:
\[ \gamma + \alpha = -\beta \]
\[ \alpha + \beta = -\gamma \]

4. **Calculate the Summation**:
We need to find:
\[ \sum \frac{1}{\beta + \gamma} = \frac{1}{\beta + \gamma} + \frac{1}{\gamma + \alpha} + \frac{1}{\alpha + \beta} \]
Substituting the expressions for \(\beta + \gamma\), \(\gamma + \alpha\), and \(\alpha + \beta\):
\[ = \frac{1}{-\alpha} + \frac{1}{-\beta} + \frac{1}{-\gamma} \]
\[ = -\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right) \]

5. **Combine the Terms**:
The expression \(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\) can be rewritten using the product of the roots:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \]
Substituting the values from Vieta's formulas:
\[ = \frac{q}{-r} = -\frac{q}{r} \]

6. **Final Result**:
Therefore, we have:
\[ \sum \frac{1}{\beta + \gamma} = -\left(-\frac{q}{r}\right) = \frac{q}{r} \]

### Conclusion:
The value of \(\sum (\beta + \gamma)^{-1}\) is \(\frac{q}{r}\).