`sum (1)/(alpha^2 beta^2) `

Options

A. `(q^2 -2pr)/(r^2)`
B. `q^3 -3pqr + 3r^2`
C. `(p^2 - 2q)/( r^2)`
D. ` (pq)/( r-3)`

Answer

To solve the given problem, we need to find the expression for the summation \( \sum \frac{1}{\alpha^2 \beta^2} \), where \( \alpha, \beta, \gamma \) are the roots of a cubic equation. Let's go through the steps systematically.

### Step-by-Step Solution

1. **Understanding the Expression**:
The expression we need to evaluate is \( \sum \frac{1}{\alpha^2 \beta^2} \). This can be rewritten as:
\[ \sum \frac{1}{\alpha^2 \beta^2} = \frac{1}{\alpha^2 \beta^2} + \frac{1}{\beta^2 \gamma^2} + \frac{1}{\gamma^2 \alpha^2} \]

2. **Finding a Common Denominator**:
The common denominator for the fractions is \( \alpha^2 \beta^2 \gamma^2 \). Thus, we can express the summation as:
\[ \sum \frac{1}{\alpha^2 \beta^2} = \frac{\gamma^2 + \alpha^2 + \beta^2}{\alpha^2 \beta^2 \gamma^2} \]

3. **Using the Relationship Between Roots and Coefficients**:
For a cubic equation \( x^3 + p_1 x^2 + p_2 x + p_3 = 0 \), we have the following relationships:
- \( \alpha + \beta + \gamma = -p_1 \)
- \( \alpha \beta + \beta \gamma + \gamma \alpha = p_2 \)
- \( \alpha \beta \gamma = -p_3 \)

4. **Calculating \( \alpha^2 + \beta^2 + \gamma^2 \)**:
We can express \( \alpha^2 + \beta^2 + \gamma^2 \) using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha) \]
Substituting the relationships:
\[ \alpha^2 + \beta^2 + \gamma^2 = (-p_1)^2 - 2p_2 = p_1^2 - 2p_2 \]

5. **Substituting Back into the Summation**:
Now substituting \( \alpha^2 + \beta^2 + \gamma^2 \) back into our expression:
\[ \sum \frac{1}{\alpha^2 \beta^2} = \frac{p_1^2 - 2p_2}{\alpha^2 \beta^2 \gamma^2} \]

6. **Expressing \( \alpha^2 \beta^2 \gamma^2 \)**:
We know \( \alpha^2 \beta^2 \gamma^2 = (\alpha \beta \gamma)^2 = (-p_3)^2 = p_3^2 \).

7. **Final Expression**:
Thus, we can write:
\[ \sum \frac{1}{\alpha^2 \beta^2} = \frac{p_1^2 - 2p_2}{p_3^2} \]

### Final Answer
The final result for the summation \( \sum \frac{1}{\alpha^2 \beta^2} \) is:
\[ \sum \frac{1}{\alpha^2 \beta^2} = \frac{p_1^2 - 2p_2}{p_3^2} \]