If `alpha, beta and gamma` are the roots of the equation `x^(3)-px^(2)+qx-r=0`, then the value of `(alphabeta)/(gamma)+(betagamma)/(alpha)+(gammaalpha)/(beta)` is equal to

Options

A. `pq+3r`
B. `pq+r`
C. `pq-3r`
D. `(q^(2)-2pr)/(r )`

Answer

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

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

where \(\alpha\), \(\beta\), and \(\gamma\) are the roots of the polynomial equation:

\[ x^3 - px^2 + qx - r = 0 \]

### Step 1: Use Vieta's Formulas
From Vieta's formulas, we know the following relationships for the roots of the polynomial:

1. \(\alpha + \beta + \gamma = p\)
2. \(\alpha \beta + \beta \gamma + \gamma \alpha = q\)
3. \(\alpha \beta \gamma = r\)

### Step 2: Rewrite the Expression
We can rewrite the expression we want to find:

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

### Step 3: Find the Numerator
We need to find \(\alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2\). This can be expressed in terms of \(q\) and \(r\):

\[ \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 = (\alpha \beta + \beta \gamma + \gamma \alpha)^2 - 2\alpha \beta \gamma (\alpha + \beta + \gamma) \]

Substituting the values from Vieta's formulas:

\[ = q^2 - 2r \cdot p \]

### Step 4: Substitute Back into the Expression
Now, substituting back into our rewritten expression:

\[ \frac{\alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2}{\alpha \beta \gamma} = \frac{q^2 - 2rp}{r} \]

### Step 5: Final Result
Thus, the final value of the expression is:

\[ \frac{q^2 - 2rp}{r} \]

### Conclusion
The value of \(\frac{\alpha \beta}{\gamma} + \frac{\beta \gamma}{\alpha} + \frac{\gamma \alpha}{\beta}\) is:

\[ \frac{q^2 - 2rp}{r} \]