If ` alpha , beta , gamma ` are the roots of ` x^3 +px^2 +qx +r=0` then find
` sum alpha^2 beta + sum alpha beta ^2`

Answer

To solve the problem, we need to find the value of \( \sum \alpha^2 \beta + \sum \alpha \beta^2 \) given that \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 + px^2 + qx + r = 0 \).

### Step-by-Step Solution:

1. **Identify the roots and coefficients**:
The polynomial is given as \( x^3 + px^2 + qx + r = 0 \). The roots are \( \alpha, \beta, \gamma \).
- From Vieta's formulas, we have:
- \( \alpha + \beta + \gamma = -p \)
- \( \alpha\beta + \beta\gamma + \gamma\alpha = q \)
- \( \alpha\beta\gamma = -r \)

2. **Express the required sum**:
We need to find \( \sum \alpha^2 \beta + \sum \alpha \beta^2 \). This can be expanded as:
\[ \alpha^2 \beta + \beta^2 \alpha + \beta^2 \gamma + \gamma^2 \beta + \gamma^2 \alpha + \alpha^2 \gamma \]

3. **Rearranging the terms**:
We can rearrange the expression:
\[ \sum \alpha^2 \beta + \sum \alpha \beta^2 = \alpha^2 \beta + \beta^2 \alpha + \beta^2 \gamma + \gamma^2 \beta + \gamma^2 \alpha + \alpha^2 \gamma \]
This can be grouped as:
\[ = \alpha \beta (\alpha + \beta) + \beta \gamma (\beta + \gamma) + \gamma \alpha (\gamma + \alpha) \]

4. **Substituting the values**:
Using \( \alpha + \beta + \gamma = -p \), we can substitute:
\[ = \alpha \beta (-p - \gamma) + \beta \gamma (-p - \alpha) + \gamma \alpha (-p - \beta) \]
Expanding this gives:
\[ = -p(\alpha \beta + \beta \gamma + \gamma \alpha) - (\alpha \beta \gamma + \beta \gamma \alpha + \gamma \alpha \beta) \]

5. **Using Vieta's relations**:
Substitute \( \alpha \beta + \beta \gamma + \gamma \alpha = q \) and \( \alpha \beta \gamma = -r \):
\[ = -pq - (-r) = -pq + r \]

6. **Final result**:
Thus, the required value is:
\[ \sum \alpha^2 \beta + \sum \alpha \beta^2 = -pq + 3r \]

### Final Answer:
\[ \sum \alpha^2 \beta + \sum \alpha \beta^2 = 3r - pq \]