If `alpha , beta , gamma ` are the roots of the equation `x^3 -6x^2 +11 x +6=0` then ` sum alpha^2 beta =`

Options

A. 80
B. 94
C. 90
D. `-84`

Answer

To find the value of \( \sigma \alpha^2 \beta \) where \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 - 6x^2 + 11x + 6 = 0 \), we will follow these steps:

### Step 1: Identify the coefficients and roots
The given polynomial is \( x^3 - 6x^2 + 11x + 6 = 0 \). According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta + \gamma = 6 \) (coefficient of \( x^2 \) with a negative sign).
- The sum of the product of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = 11 \) (coefficient of \( x \)).
- The product of the roots \( \alpha \beta \gamma = -6 \) (constant term with a negative sign).

### Step 2: Express \( \sigma \alpha^2 \beta \)
We need to find \( \sigma \alpha^2 \beta = \alpha^2 \beta + \alpha \beta^2 + \beta^2 \gamma + \beta \gamma^2 + \gamma^2 \alpha + \gamma \alpha^2 \).

### Step 3: Group terms and factor
We can group the terms as follows:
\[ \sigma \alpha^2 \beta = \alpha \beta (\alpha + \beta) + \beta \gamma (\beta + \gamma) + \gamma \alpha (\gamma + \alpha) \]
This can be rewritten as:
\[ = \alpha \beta (6 - \gamma) + \beta \gamma (6 - \alpha) + \gamma \alpha (6 - \beta) \]

### Step 4: Substitute values
Now substituting \( \alpha + \beta = 6 - \gamma \), \( \beta + \gamma = 6 - \alpha \), and \( \gamma + \alpha = 6 - \beta \):
\[ = \alpha \beta (6 - \gamma) + \beta \gamma (6 - \alpha) + \gamma \alpha (6 - \beta) \]
Expanding this gives:
\[ = 6(\alpha \beta + \beta \gamma + \gamma \alpha) - (\alpha \beta \gamma + \beta \gamma \alpha + \gamma \alpha \beta) \]

### Step 5: Substitute known values
We know:
- \( \alpha \beta + \beta \gamma + \gamma \alpha = 11 \)
- \( \alpha \beta \gamma = -6 \)

Substituting these values:
\[ = 6(11) - 3(\alpha \beta \gamma) = 66 - 3(-6) = 66 + 18 = 84 \]

### Final Answer
Thus, the value of \( \sigma \alpha^2 \beta \) is \( 84 \).

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