If ` alpha , beta ` are the roots of ` ax^(2) + bx +c=0` , then ` (alpha^(3) + beta^(3))/(alpha^(-3) + beta^(-3))` is equal to :

Options

A. `(c^(2))/(a^(2))`
B. `(c^(3))/(a^(3))`
C. `(3abc-b^(3))/(c^(3))`
D. `(b^(2)-2ac)/(ac)`

Answer

To solve the problem, we need to find the value of \(\frac{\alpha^3 + \beta^3}{\alpha^{-3} + \beta^{-3}}\) where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(ax^2 + bx + c = 0\).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ \frac{\alpha^3 + \beta^3}{\alpha^{-3} + \beta^{-3}} \]

2. **Simplifying the Denominator**:
The denominator \(\alpha^{-3} + \beta^{-3}\) can be rewritten using the property of exponents:
\[ \alpha^{-3} + \beta^{-3} = \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\beta^3 + \alpha^3}{\alpha^3 \beta^3} \]
Therefore, we can rewrite the expression as:
\[ \frac{\alpha^3 + \beta^3}{\frac{\alpha^3 + \beta^3}{\alpha^3 \beta^3}} = \frac{(\alpha^3 + \beta^3) \cdot \alpha^3 \beta^3}{\alpha^3 + \beta^3} \]

3. **Cancelling the Common Terms**:
Since \(\alpha^3 + \beta^3\) is in both the numerator and the denominator, we can cancel it (assuming \(\alpha^3 + \beta^3 \neq 0\)):
\[ = \alpha^3 \beta^3 \]

4. **Using Vieta's Formulas**:
From Vieta's formulas, we know that:
\[ \alpha \beta = \frac{c}{a} \]
Therefore,
\[ \alpha^3 \beta^3 = (\alpha \beta)^3 = \left(\frac{c}{a}\right)^3 = \frac{c^3}{a^3} \]

5. **Final Result**:
Thus, we conclude that:
\[ \frac{\alpha^3 + \beta^3}{\alpha^{-3} + \beta^{-3}} = \frac{c^3}{a^3} \]

### Answer:
The final answer is:
\[ \frac{c^3}{a^3} \]