If `alpha, beta and gamma` are the zeroes of the polynomial `f(x) = ax^(3) + bx^(2) + cx + d`, then `(1)/(alpha) +(1)/( beta) + (1)/(gamma) = `

Options

A. `- (d)/( b)`
B. `( c)/( d)`
C. `- ( c)/( d)`
D. ` - ( c)/( a)`

Answer

To solve the problem, we need to find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \) given that \( \alpha, \beta, \) and \( \gamma \) are the roots of the polynomial \( f(x) = ax^3 + bx^2 + cx + d \).

### Step-by-Step Solution:

1. **Understanding the Roots:**
The roots of the polynomial are given as \( \alpha, \beta, \gamma \). According to Vieta's formulas, we can express the sums and products of the roots in terms of the coefficients of the polynomial.

2. **Sum of Roots:**
The sum of the roots \( \alpha + \beta + \gamma \) can be given by:
\[ \alpha + \beta + \gamma = -\frac{b}{a} \]

3. **Sum of Product of Roots Taken Two at a Time:**
The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha \) is given by:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \]

4. **Product of Roots:**
The product of the roots \( \alpha\beta\gamma \) is given by:
\[ \alpha\beta\gamma = -\frac{d}{a} \]

5. **Finding \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \):**
We can rewrite \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \) as:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \]
Substituting the values from Vieta's formulas:
\[ = \frac{\frac{c}{a}}{-\frac{d}{a}} = -\frac{c}{d} \]

6. **Final Result:**
Therefore, the value of \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \) is:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = -\frac{c}{d} \]