If α,β are the roots of `x^(2)-x+2=0` then α^3 β+αβ^3

Options

A. a. -6
B. b. 6
C. c. -10
D. d. all of the abvoe

Answer

To solve the problem, we need to find the value of \( \alpha^3 \beta + \alpha \beta^3 \) given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - x + 2 = 0 \).

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation is \( x^2 - x + 2 = 0 \). Here, we have:
- \( a = 1 \)
- \( b = -1 \)
- \( c = 2 \)

2. **Find the sum and product of the roots**:
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{2}{1} = 2 \)

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

4. **Find \( \alpha^2 + \beta^2 \)**:
We can use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = (1)^2 - 2(2) = 1 - 4 = -3 \]

5. **Substitute back into the expression**:
Now we substitute \( \alpha \beta \) and \( \alpha^2 + \beta^2 \) back into our expression:
\[ \alpha^3 \beta + \alpha \beta^3 = \alpha \beta (\alpha^2 + \beta^2) = 2 \cdot (-3) = -6 \]

### Final Answer:
Thus, the value of \( \alpha^3 \beta + \alpha \beta^3 \) is \( -6 \).