Find the values of `alpha^(3)+beta^(3)` in terms of a, b, c if `alpha, beta` are roots of `ax^(2)+bx+c=0`, `c != 0`

Answer

To find the value of \( \alpha^3 + \beta^3 \) in terms of \( a, b, c \) where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( ax^2 + bx + c = 0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Use the identity for the sum of cubes**:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We can rewrite \( \alpha^2 - \alpha\beta + \beta^2 \) as \( (\alpha + \beta)^2 - 3\alpha\beta \).

2. **Find \( \alpha + \beta \) and \( \alpha\beta \)**:
From Vieta's formulas, we know:
\[ \alpha + \beta = -\frac{b}{a} \]
\[ \alpha\beta = \frac{c}{a} \]

3. **Substitute these values into the identity**:
Now, substituting \( \alpha + \beta \) and \( \alpha\beta \) into our expression:
\[ \alpha^3 + \beta^3 = \left(-\frac{b}{a}\right) \left(\left(-\frac{b}{a}\right)^2 - 3\left(\frac{c}{a}\right)\right) \]

4. **Simplify the expression**:
First, calculate \( \left(-\frac{b}{a}\right)^2 \):
\[ \left(-\frac{b}{a}\right)^2 = \frac{b^2}{a^2} \]
Now, substitute this back into the equation:
\[ \alpha^3 + \beta^3 = \left(-\frac{b}{a}\right) \left(\frac{b^2}{a^2} - 3\frac{c}{a}\right) \]
This simplifies to:
\[ \alpha^3 + \beta^3 = -\frac{b}{a} \left(\frac{b^2 - 3ac}{a^2}\right) \]

5. **Combine the fractions**:
\[ \alpha^3 + \beta^3 = -\frac{b(b^2 - 3ac)}{a^3} \]

6. **Final result**:
Therefore, the final expression for \( \alpha^3 + \beta^3 \) in terms of \( a, b, c \) is:
\[ \alpha^3 + \beta^3 = \frac{-b^3 + 3abc}{a^3} \]