If `alpha, beta` are roots of the equations `Ax^(2)+Bx+C=0`. Then value of `alpha^(3)+beta^(3)` is

Options

A. `(3ABC-B^(3))/(A^(3))`
B. `(3ABC +B^(3))/(A^(3))`
C. `(B^(3)-3ABC)/(A^(3))`
D. none of these

Answer

To find the value of \( \alpha^3 + \beta^3 \) where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( Ax^2 + Bx + C = 0 \), we can use the relationships derived from Vieta's formulas.

### Step-by-Step Solution:

1. **Identify the Sum and Product of Roots:**
From Vieta's formulas, we know:
\[ \alpha + \beta = -\frac{B}{A} \]
\[ \alpha \beta = \frac{C}{A} \]

2. **Use the Identity for Cubes:**
We can express \( \alpha^3 + \beta^3 \) using the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We can rewrite \( \alpha^2 + \beta^2 \) using the square of the sum of the roots:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Thus,
\[ \alpha^2 - \alpha\beta + \beta^2 = (\alpha^2 + \beta^2) - \alpha\beta = \left((\alpha + \beta)^2 - 2\alpha\beta\right) - \alpha\beta \]
\[ = (\alpha + \beta)^2 - 3\alpha\beta \]

3. **Substitute the Values:**
Substitute \( \alpha + \beta \) and \( \alpha \beta \) into the equation:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)\left((\alpha + \beta)^2 - 3\alpha\beta\right) \]
\[ = \left(-\frac{B}{A}\right)\left(\left(-\frac{B}{A}\right)^2 - 3\left(\frac{C}{A}\right)\right) \]

4. **Simplify the Expression:**
Now we simplify:
\[ = -\frac{B}{A}\left(\frac{B^2}{A^2} - \frac{3C}{A}\right) \]
\[ = -\frac{B}{A}\left(\frac{B^2 - 3AC}{A^2}\right) \]
\[ = -\frac{B(B^2 - 3AC)}{A^3} \]

5. **Final Result:**
Therefore, the value of \( \alpha^3 + \beta^3 \) is:
\[ \alpha^3 + \beta^3 = \frac{-B^3 + 3ABC}{A^3} \]