If ` alpha , beta , gamma ` are roots of the equation `x^3 + ax^2 + bx +c=0` then ` alpha^(-1) + beta^(-1) + gamma^(-1) =`

Options

A. `a/c`
B. `-b/c`
C. `c/a`
D. `b/a`

Answer

To solve the problem, we need to find the expression \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \) given that \( \alpha, \beta, \gamma \) are the roots of the polynomial equation \( x^3 + ax^2 + bx + c = 0 \).

### Step-by-step Solution:

1. **Understanding the Roots**:
The roots of the polynomial \( x^3 + ax^2 + bx + c = 0 \) are \( \alpha, \beta, \gamma \).

2. **Using Vieta's Formulas**:
According to Vieta's formulas for a cubic polynomial:
- The sum of the roots \( \alpha + \beta + \gamma = -a \).
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = b \).
- The product of the roots \( \alpha\beta\gamma = -c \).

3. **Finding the Expression**:
We want to find \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \). This can be rewritten using the common denominator:
\[ \alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \]

4. **Substituting Values**:
From Vieta's formulas:
- The numerator \( \beta\gamma + \gamma\alpha + \alpha\beta = b \).
- The denominator \( \alpha\beta\gamma = -c \).

Therefore, we can substitute these values into the expression:
\[ \alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{b}{-c} = -\frac{b}{c} \]

5. **Final Result**:
Thus, the final result is:
\[ \alpha^{-1} + \beta^{-1} + \gamma^{-1} = -\frac{b}{c} \]

### Answer:
The answer is \( -\frac{b}{c} \).

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