If `alpha,beta` are the roots of `ax^(2)+bx+c=0`, find the value of
(i) `((alpha)/(beta)-(beta)/(alpha))^(2)`
(ii) `(alpha^(3))/(beta)+(beta^(3))/(alpha)`.

Answer

To solve the given problem, we need to find the values of two expressions involving the roots \( \alpha \) and \( \beta \) of the quadratic equation \( ax^2 + bx + c = 0 \).

### Step-by-step Solution:

**Given:**
- The roots of the equation \( ax^2 + bx + c = 0 \) are \( \alpha \) and \( \beta \).
- From Vieta's formulas, we know:
- \( \alpha + \beta = -\frac{b}{a} \)
- \( \alpha \beta = \frac{c}{a} \)

#### Part (i): Find \( \left( \frac{\alpha}{\beta} - \frac{\beta}{\alpha} \right)^2 \)

1. **Simplify the expression:**
\[ \frac{\alpha}{\beta} - \frac{\beta}{\alpha} = \frac{\alpha^2 - \beta^2}{\alpha \beta} \]
Here, we find a common denominator.

2. **Factor \( \alpha^2 - \beta^2 \):**
\[ \alpha^2 - \beta^2 = (\alpha - \beta)(\alpha + \beta) \]
Therefore,
\[ \frac{\alpha}{\beta} - \frac{\beta}{\alpha} = \frac{(\alpha - \beta)(\alpha + \beta)}{\alpha \beta} \]

3. **Substituting values from Vieta's formulas:**
- \( \alpha + \beta = -\frac{b}{a} \)
- \( \alpha \beta = \frac{c}{a} \)

Thus,
\[ \frac{\alpha}{\beta} - \frac{\beta}{\alpha} = \frac{(\alpha - \beta)(-\frac{b}{a})}{\frac{c}{a}} = \frac{(\alpha - \beta)(-b)}{c} \]

4. **Square the expression:**
\[ \left( \frac{\alpha}{\beta} - \frac{\beta}{\alpha} \right)^2 = \left( \frac{(\alpha - \beta)(-b)}{c} \right)^2 = \frac{(\alpha - \beta)^2 b^2}{c^2} \]

5. **Using the identity for \( (\alpha - \beta)^2 \):**
\[ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \]
Substitute the values:
\[ (\alpha - \beta)^2 = \left(-\frac{b}{a}\right)^2 - 4 \cdot \frac{c}{a} = \frac{b^2}{a^2} - \frac{4c}{a} \]
Multiply by \( a^2 \):
\[ = \frac{b^2 - 4ac}{a^2} \]

6. **Final expression for part (i):**
\[ \left( \frac{\alpha}{\beta} - \frac{\beta}{\alpha} \right)^2 = \frac{(b^2 - 4ac)b^2}{c^2 a^2} \]

#### Part (ii): Find \( \frac{\alpha^3}{\beta} + \frac{\beta^3}{\alpha} \)

1. **Combine the fractions:**
\[ \frac{\alpha^3}{\beta} + \frac{\beta^3}{\alpha} = \frac{\alpha^4 + \beta^4}{\alpha \beta} \]

2. **Use the identity for \( \alpha^4 + \beta^4 \):**
\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \]

3. **Calculate \( \alpha^2 + \beta^2 \):**
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{b}{a}\right)^2 - 2\cdot\frac{c}{a} = \frac{b^2}{a^2} - \frac{2c}{a} \]

4. **Substituting back:**
\[ \alpha^4 + \beta^4 = \left( \frac{b^2 - 2ac}{a^2} \right)^2 - 2\left( \frac{c}{a} \right)^2 \]

5. **Final expression for part (ii):**
\[ \frac{\alpha^3}{\beta} + \frac{\beta^3}{\alpha} = \frac{(\alpha^4 + \beta^4)}{\alpha \beta} = \frac{\left( \frac{(b^2 - 2ac)^2 - 2c^2}{a^4} \right)}{\frac{c}{a}} = \frac{(b^2 - 2ac)^2 - 2c^2}{ac} \]