If (alpha) and(beta) be the roots of the equation `ax^2+bx + c=0`, find the value of
`(alpha^(2))/(beta) + (beta^(2))/(alpha)`

Options

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

Answer

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

### Step-by-step Solution:

1. **Use Vieta's Formulas**:
From Vieta's formulas, we know that:
- \(\alpha + \beta = -\frac{b}{a}\) (sum of roots)
- \(\alpha \beta = \frac{c}{a}\) (product of roots)

2. **Rewrite the Expression**:
We need to simplify the expression \(\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}\). We can rewrite it as:
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha \beta} \]

3. **Use the Identity for Sum of Cubes**:
The sum of cubes can be expressed as:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We can also express \(\alpha^2 + \beta^2\) using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Thus,
\[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) \]

4. **Substitute Vieta's Values**:
Now substituting the values from Vieta's formulas:
\[ \alpha^3 + \beta^3 = \left(-\frac{b}{a}\right)\left(\left(-\frac{b}{a}\right)^2 - 3\frac{c}{a}\right) \]
Simplifying this gives:
\[ = -\frac{b}{a}\left(\frac{b^2}{a^2} - \frac{3c}{a}\right) = -\frac{b}{a}\left(\frac{b^2 - 3ac}{a^2}\right) = -\frac{b(b^2 - 3ac)}{a^3} \]

5. **Combine Everything**:
Now substituting back into our expression:
\[ \frac{\alpha^3 + \beta^3}{\alpha \beta} = \frac{-\frac{b(b^2 - 3ac)}{a^3}}{\frac{c}{a}} = -\frac{b(b^2 - 3ac)}{ac} \]

6. **Final Expression**:
Thus, we have:
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{3ac - b^2}{ac} \]

### Conclusion:
The final value of \(\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}\) is:
\[ \frac{3ac - b^2}{ac} \]