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 equation \(x^2 + x + 1 = 0\) and then evaluate the expressions involving these roots. Let's go through the solution step by step.

### Step 1: Find the roots of the equation \(x^2 + x + 1 = 0\)

We can use the quadratic formula to find the roots of the equation. The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation, \(a = 1\), \(b = 1\), and \(c = 1\). Plugging in these values, we get:

\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]

### Step 2: Calculate the discriminant

Now, calculate the discriminant:

\[ b^2 - 4ac = 1 - 4 = -3 \]

Since the discriminant is negative, the roots will be complex.

### Step 3: Substitute the discriminant back into the formula

Substituting the discriminant back into the formula gives us:

\[ x = \frac{-1 \pm \sqrt{-3}}{2} \]

### Step 4: Simplify the roots

We can express \(\sqrt{-3}\) as \(i\sqrt{3}\):

\[ x = \frac{-1 \pm i\sqrt{3}}{2} \]

Thus, the roots are:

\[ \alpha = \frac{-1 + i\sqrt{3}}{2}, \quad \beta = \frac{-1 - i\sqrt{3}}{2} \]

### Step 5: Identify the roots in terms of cube roots of unity

The roots \(\alpha\) and \(\beta\) can be recognized as the cube roots of unity:

\[ \alpha = \omega, \quad \beta = \omega^2 \]

where \(\omega = e^{2\pi i / 3} = \frac{-1 + i\sqrt{3}}{2}\) and \(\omega^2 = e^{-2\pi i / 3} = \frac{-1 - i\sqrt{3}}{2}\).

### Step 6: Evaluate \( \alpha^3 + \beta^3 \)

Using the property of cube roots of unity, we know:

\[ \alpha^3 = 1, \quad \beta^3 = 1 \]

Thus:

\[ \alpha^3 + \beta^3 = 1 + 1 = 2 \]

### Conclusion

The correct option is that \( \alpha^3 + \beta^3 = 2 \).