If ` alpha , beta ` are the roots of ` ax ^2 + bx +c=0` then `alpha ^5 beta ^8 + alpha^8 beta ^5=`

Options

A. `(c^(5)(3abc-b^(3))/(a^(8)))`
B. `(c^(3)(3abc-b^(3))/(a^(5)))`
C. `(a^(2)(3abc-b^(3))/(c^(5)))`
D. `(a^(3)(3abc-b^(3))/(c^(5)))`

Answer

To solve the problem, we need to find the value of \( \alpha^5 \beta^8 + \alpha^8 \beta^5 \) given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( ax^2 + bx + c = 0 \).

### Step-by-Step Solution

1. **Identify the roots**:
From Vieta's formulas, we know:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a} \]

2. **Rewrite the expression**:
We can factor the expression \( \alpha^5 \beta^8 + \alpha^8 \beta^5 \):
\[ \alpha^5 \beta^8 + \alpha^8 \beta^5 = \alpha^5 \beta^5 (\beta^3 + \alpha^3) \]

3. **Find \( \beta^3 + \alpha^3 \)**:
We can use the identity for the sum of cubes:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We can express \( \alpha^2 + \beta^2 \) as:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Thus,
\[ \alpha^2 - \alpha\beta + \beta^2 = \alpha^2 + \beta^2 - \alpha\beta = \left( (\alpha + \beta)^2 - 2\alpha\beta \right) - \alpha\beta \]
Substituting the values:
\[ = \left( -\frac{b}{a} \right)^2 - 3 \cdot \frac{c}{a} = \frac{b^2}{a^2} - \frac{3c}{a} \]

4. **Substituting back**:
Now substituting back into the expression for \( \alpha^3 + \beta^3 \):
\[ \alpha^3 + \beta^3 = \left( -\frac{b}{a} \right) \left( \frac{b^2}{a^2} - \frac{3c}{a} \right) = -\frac{b}{a} \cdot \frac{b^2 - 3ac}{a^2} = -\frac{b(b^2 - 3ac)}{a^3} \]

5. **Combine everything**:
Now we can substitute back into our expression:
\[ \alpha^5 \beta^5 (\beta^3 + \alpha^3) = \left( \frac{c}{a} \right)^5 \left( -\frac{b(b^2 - 3ac)}{a^3} \right) \]
This simplifies to:
\[ = -\frac{c^5 b (b^2 - 3ac)}{a^8} \]

### Final Answer
Thus, the final expression for \( \alpha^5 \beta^8 + \alpha^8 \beta^5 \) is:
\[ -\frac{c^5 b (b^2 - 3ac)}{a^8} \]