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

Options

A. `p^2 -2q`
B. `p^3 -3pq +3r`
C. `p^4 -3p^2 q +2q^2 `
D. `2q`

Answer

To find the value of \( \alpha^2 + \beta^2 + \gamma^2 \) given that \( \alpha, \beta, \gamma \) are the roots of the polynomial equation \( x^3 + px^2 + qx - r = 0 \), we can use the relationships derived from Vieta's formulas.

### Step-by-step Solution:

1. **Identify the coefficients from the polynomial:**
The given polynomial is \( x^3 + px^2 + qx - r = 0 \). From this, we can identify:
- Coefficient of \( x^2 \) is \( p \)
- Coefficient of \( x \) is \( q \)
- Constant term is \( -r \)

2. **Use Vieta's Formulas:**
According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -p \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \alpha\gamma = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} = q \)
- The product of the roots \( \alpha\beta\gamma = -\frac{\text{constant term}}{\text{coefficient of } x^3} = r \)

3. **Find \( \alpha^2 + \beta^2 + \gamma^2 \):**
We can express \( \alpha^2 + \beta^2 + \gamma^2 \) using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \alpha\gamma) \]
Substituting the values from Vieta's formulas:
\[ \alpha^2 + \beta^2 + \gamma^2 = (-p)^2 - 2q \]

4. **Simplify the expression:**
\[ \alpha^2 + \beta^2 + \gamma^2 = p^2 - 2q \]

Thus, the final result is:
\[ \alpha^2 + \beta^2 + \gamma^2 = p^2 - 2q \]