If ` alpha , beta , gamma ` are the roots of ` x^3 +px^2 +qx +r=0` then find
`sum (1)/( alpha )`

Answer

To find the value of the summation \( \sum \frac{1}{\alpha} \) where \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 + px^2 + qx + r = 0 \), we can follow these steps:

### Step 1: Understand the roots and the polynomial
The given polynomial is \( x^3 + px^2 + qx + r = 0 \). The roots of this polynomial are \( \alpha, \beta, \gamma \).

### Step 2: Use Vieta's formulas
According to Vieta's formulas for a cubic equation \( x^3 + ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -p \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = q \)
- The product of the roots \( \alpha\beta\gamma = -r \)

### Step 3: Write the expression for \( \sum \frac{1}{\alpha} \)
The expression \( \sum \frac{1}{\alpha} \) can be rewritten as:
\[ \sum \frac{1}{\alpha} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \]
This can be combined into a single fraction:
\[ \sum \frac{1}{\alpha} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma} \]

### Step 4: Substitute the values from Vieta's formulas
From Vieta's formulas:
- The numerator \( \beta\gamma + \alpha\gamma + \alpha\beta = q \)
- The denominator \( \alpha\beta\gamma = -r \)

Thus, we can substitute these values into the expression:
\[ \sum \frac{1}{\alpha} = \frac{q}{-\r} \]

### Step 5: Final result
Therefore, the value of \( \sum \frac{1}{\alpha} \) is:
\[ \sum \frac{1}{\alpha} = -\frac{q}{r} \]