If ` alpha , beta , gamma` are the roots of ` x^3 -px^2 +qx -r=0` and `r ne 0` then find `(1)/( alpha^2) +(1)/( beta^2) +(1)/( gamma ^2)` in terms of p,q ,r

Answer

To solve the problem, we need to find the value of \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} \) in terms of \( p, q, \) and \( r \), 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 their relationships**:
The roots of the polynomial \( x^3 - px^2 + qx - r = 0 \) are \( \alpha, \beta, \gamma \). By Vieta's formulas, we know:
- \( \alpha + \beta + \gamma = p \)
- \( \alpha\beta + \beta\gamma + \gamma\alpha = q \)
- \( \alpha\beta\gamma = r \)

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

3. **Calculate the numerator**:
The numerator \( \beta^2\gamma^2 + \alpha^2\gamma^2 + \alpha^2\beta^2 \) can be expressed as:
\[ \beta^2\gamma^2 + \alpha^2\gamma^2 + \alpha^2\beta^2 = (\alpha\beta)^2 + (\beta\gamma)^2 + (\gamma\alpha)^2 \]
We can use the identity:
\[ (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \]
where \( x = \alpha\beta, y = \beta\gamma, z = \gamma\alpha \).

4. **Relate the terms**:
From Vieta's formulas, we know:
\[ xy + yz + zx = \alpha\beta\gamma(\alpha + \beta + \gamma) = r \cdot p \]
Thus,
\[ (\alpha\beta + \beta\gamma + \gamma\alpha)^2 = q^2 \]
Therefore,
\[ \beta^2\gamma^2 + \alpha^2\gamma^2 + \alpha^2\beta^2 = q^2 - 2r \cdot p \]

5. **Calculate the denominator**:
The denominator \( \alpha^2\beta^2\gamma^2 \) can be expressed as:
\[ (\alpha\beta\gamma)^2 = r^2 \]

6. **Combine the results**:
Now substituting back into our expression, we have:
\[ \frac{\beta^2\gamma^2 + \alpha^2\gamma^2 + \alpha^2\beta^2}{\alpha^2\beta^2\gamma^2} = \frac{q^2 - 2rp}{r^2} \]

### Final Answer:
Thus, the final result is:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{q^2 - 2rp}{r^2} \]