If `alpha , beta , gamma` be the roots of the equation
`x ( 1+ x^(2)) + x^(2) ( 6 + x) + 2 = 0` , then the value of `alpha^(-1) + beta^(-1) + gamma^(-1)`
A)`-3`
B)`1/2`
C)`-1/2`
D)None of these

Options

A. `-3`
B. `1/2`
C. `-1/2`
D. None of these

Answer

To solve the problem, we need to find the value of \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \) where \( \alpha, \beta, \gamma \) are the roots of the cubic equation given by:

\[ x(1 + x^2) + x^2(6 + x) + 2 = 0 \]

### Step 1: Expand the equation
First, we will expand the equation to obtain a standard cubic form.

\[ x(1 + x^2) = x + x^3 \]
\[ x^2(6 + x) = 6x^2 + x^3 \]

Combining these, we have:

\[ x + x^3 + 6x^2 + x^3 + 2 = 0 \]

This simplifies to:

\[ 2x^3 + 6x^2 + x + 2 = 0 \]

### Step 2: Identify coefficients
Now, we can identify the coefficients of the cubic equation \( ax^3 + bx^2 + cx + d = 0 \):

- \( a = 2 \)
- \( b = 6 \)
- \( c = 1 \)
- \( d = 2 \)

### Step 3: Use Vieta's formulas
According to Vieta's formulas for a cubic equation, we have the following relationships for the roots \( \alpha, \beta, \gamma \):

1. \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{6}{2} = -3 \)
2. \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{1}{2} \)
3. \( \alpha\beta\gamma = -\frac{d}{a} = -\frac{2}{2} = -1 \)

### Step 4: Calculate \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \)
We can express \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \) in terms of the roots:

\[ \alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \]

Substituting the values from Vieta's formulas:

\[ \alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{\frac{1}{2}}{-1} = -\frac{1}{2} \]

### Conclusion
Thus, the value of \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \) is:

\[ \boxed{-\frac{1}{2}} \]