Let ` alpha, and beta` are the roots of the equation ` x^(2)+x +1 =0` then

Options

A. `alpha^(2) +beta^(2) =4`
B. `(alpha - beta)^(2)=3`
C. `alpha^(3) +beta^(3)=2`
D. `alpha^(4) +beta^(4) = 1`

Answer

To solve the problem, we need to find the roots of the quadratic equation \( x^2 + x + 1 = 0 \) and then evaluate various expressions involving these roots, denoted as \( \alpha \) and \( \beta \).

### Step 1: Find the roots of the equation
The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \( x^2 + x + 1 = 0 \), we have:
- \( a = 1 \)
- \( b = 1 \)
- \( c = 1 \)

Substituting these values into the formula:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ x = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, the roots are:
\[ \alpha = \frac{-1 + i\sqrt{3}}{2}, \quad \beta = \frac{-1 - i\sqrt{3}}{2} \]

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

### Step 3: Calculate \( \alpha^2 + \beta^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = (-1)^2 - 2 \cdot 1 = 1 - 2 = -1 \]

### Step 4: Calculate \( \alpha - \beta \)
Using the identity:
\[ \alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} \]
Substituting the values:
\[ \alpha - \beta = \sqrt{(-1)^2 - 4 \cdot 1} = \sqrt{1 - 4} = \sqrt{-3} = i\sqrt{3} \]
Thus, \( (\alpha - \beta)^2 = -3 \).

### Step 5: Calculate \( \alpha^3 + \beta^3 \)
Using the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) \]
Substituting the values:
\[ \alpha^3 + \beta^3 = (-1)(-1 - 1) = (-1)(-2) = 2 \]

### Step 6: Calculate \( \alpha^4 + \beta^4 \)
Using the identity:
\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 \]
Substituting the values:
\[ \alpha^4 + \beta^4 = (-1)^2 - 2 \cdot (1)^2 = 1 - 2 = -1 \]

### Summary of Results
- \( \alpha + \beta = -1 \)
- \( \alpha \beta = 1 \)
- \( \alpha^2 + \beta^2 = -1 \)
- \( \alpha - \beta = i\sqrt{3} \) and \( (\alpha - \beta)^2 = -3 \)
- \( \alpha^3 + \beta^3 = 2 \)
- \( \alpha^4 + \beta^4 = -1 \)

### Final Answers
1. \( \alpha^2 + \beta^2 = -1 \) (not equal to 4)
2. \( \alpha - \beta \) squared = -3 (not equal to 3)
3. \( \alpha^3 + \beta^3 = 2 \) (correct)
4. \( \alpha^4 + \beta^4 = -1 \) (not equal to 1)

### Conclusion
The only correct option is \( \alpha^3 + \beta^3 = 2 \).