If `alpha, beta ,gamma` are roots of `x^(3) + px^(2) + qx + r = 0 ` then `sum (1)/(alpha^(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 find the value of the summation \( \sum \frac{1}{\alpha^2} \) where \( \alpha, \beta, \gamma \) are the roots of the polynomial equation \( x^3 + px^2 + qx + r = 0 \), we can follow these steps:

### Step 1: Identify the relationships between roots and coefficients
From Vieta's formulas, we know:
- The sum of the roots:
\[ \alpha + \beta + \gamma = -p \]
- The sum of the product of the roots taken two at a time:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = q \]
- The product of the roots:
\[ \alpha\beta\gamma = -r \]

### Step 2: Express the desired sum
We want to find:
\[ \sum \frac{1}{\alpha^2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} \]
This can be rewritten using a common denominator:
\[ \sum \frac{1}{\alpha^2} = \frac{\beta^2\gamma^2 + \gamma^2\alpha^2 + \alpha^2\beta^2}{\alpha^2\beta^2\gamma^2} \]

### Step 3: Find the numerator
To find \( \beta^2\gamma^2 + \gamma^2\alpha^2 + \alpha^2\beta^2 \), we can use the identity:
\[ (\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \beta^2\gamma^2 + \gamma^2\alpha^2 + \alpha^2\beta^2 + 2\alpha\beta\gamma(\alpha + \beta + \gamma) \]
Substituting the values from Vieta's formulas:
\[ q^2 = \beta^2\gamma^2 + \gamma^2\alpha^2 + \alpha^2\beta^2 + 2(-r)(-p) \]
This simplifies to:
\[ \beta^2\gamma^2 + \gamma^2\alpha^2 + \alpha^2\beta^2 = q^2 - 2rp \]

### Step 4: Find the denominator
The denominator \( \alpha^2\beta^2\gamma^2 \) can be expressed as:
\[ (\alpha\beta\gamma)^2 = (-r)^2 = r^2 \]

### Step 5: Combine results
Now substituting back into our expression for \( \sum \frac{1}{\alpha^2} \):
\[ \sum \frac{1}{\alpha^2} = \frac{q^2 - 2rp}{r^2} \]

### Final Answer
Thus, the required value is:
\[ \sum \frac{1}{\alpha^2} = \frac{q^2 - 2rp}{r^2} \]