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

Answer

To solve the problem, we need to find the value of the expression:

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

where \(\alpha\), \(\beta\), and \(\gamma\) are the roots of the cubic equation \(x^3 + qx + r = 0\).

### Step 1: Use Vieta's Formulas
From Vieta's formulas, we know:
- The sum of the roots: \(\alpha + \beta + \gamma = 0\)
- 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 2: Rewrite the Denominators
Using the relation \(\alpha + \beta + \gamma = 0\), we can rewrite the terms in the denominators:
- \(\alpha + \beta - \gamma = -\gamma\)
- \(\beta + \gamma - \alpha = -\alpha\)
- \(\gamma + \alpha - \beta = -\beta\)

Thus, we can rewrite the expression as:

\[ \frac{1}{-\gamma} + \frac{1}{-\alpha} + \frac{1}{-\beta} = -\left(\frac{1}{\gamma} + \frac{1}{\alpha} + \frac{1}{\beta}\right) \]

### Step 3: Combine the Fractions
Now, we can combine the fractions:

\[ -\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right) = -\frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \]

### Step 4: Substitute Values from Vieta's Formulas
From Vieta's, we know:
- \(\alpha\beta + \beta\gamma + \gamma\alpha = q\)
- \(\alpha\beta\gamma = -r\)

Substituting these values into the expression gives:

\[ -\frac{q}{-r} = \frac{q}{r} \]

### Final Result
Thus, the value of the expression is:

\[ \frac{q}{r} \]