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 1: Use the identity for the sum of cubes
We know that:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
Using the relationships from Vieta's formulas:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha\beta = \frac{c}{a} \]
We can express \(\alpha^2 + \beta^2\) as:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} \]
Thus:
\[ \alpha^3 + \beta^3 = \left(-\frac{b}{a}\right)\left(\frac{b^2}{a^2} - \frac{2c}{a} - \frac{c}{a}\right) = -\frac{b}{a}\left(\frac{b^2 - 2ac}{a^2}\right) = -\frac{b(b^2 - 2ac)}{a^3} \]

### Step 2: Calculate \(\alpha^{-3} + \beta^{-3}\)
We can rewrite \(\alpha^{-3} + \beta^{-3}\) as:
\[ \alpha^{-3} + \beta^{-3} = \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\alpha^3 + \beta^3}{\alpha^3 \beta^3} \]
Using the product of the roots:
\[ \alpha^3 \beta^3 = (\alpha \beta)^3 = \left(\frac{c}{a}\right)^3 = \frac{c^3}{a^3} \]
Thus:
\[ \alpha^{-3} + \beta^{-3} = \frac{\alpha^3 + \beta^3}{\frac{c^3}{a^3}} = \frac{a^3(\alpha^3 + \beta^3)}{c^3} \]

### Step 3: Substitute into the original expression
Now we substitute back into the original expression:
\[ \frac{\alpha^3 + \beta^3}{\alpha^{-3} + \beta^{-3}} = \frac{\alpha^3 + \beta^3}{\frac{a^3(\alpha^3 + \beta^3)}{c^3}} = \frac{c^3}{a^3} \]

### Final Result
Thus, we find that:
\[ \frac{\alpha^3 + \beta^3}{\alpha^{-3} + \beta^{-3}} = \frac{c^3}{a^3} \]