Let `alpha and beta` be the roots of `x^(2)+x+1=0`, then the equation whose roots are `alpha^(2020) and beta^(2020)` is

Options

A. `x^(2)+x+1=0`
B. `x^(2)-x-1=0`
C. `x^(2)+x-1=0`
D. `x^(2)-x+1=0`

Answer

To solve the problem, we need to find the equation whose roots are \( \alpha^{2020} \) and \( \beta^{2020} \), where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + x + 1 = 0 \).

### Step 1: Find the roots \( \alpha \) and \( \beta \)

The roots of the quadratic equation \( x^2 + x + 1 = 0 \) can be calculated using the quadratic formula:

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

Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \).

\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]

Thus, the roots are:

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

### Step 2: Recognize the roots as cube roots of unity

The roots \( \alpha \) and \( \beta \) can also be expressed in terms of the cube roots of unity, where:

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

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

### Step 3: Calculate \( \alpha^{2020} \) and \( \beta^{2020} \)

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

\[ \omega^3 = 1 \]

To find \( \alpha^{2020} \):

\[ 2020 \mod 3 = 1 \quad \text{(since } 2020 = 3 \cdot 673 + 1\text{)} \]

Thus,

\[ \alpha^{2020} = \omega^{2020} = \omega^1 = \omega \]

Similarly, for \( \beta^{2020} \):

\[ \beta^{2020} = (\omega^2)^{2020} = \omega^{4040} = \omega^{4040 \mod 3} = \omega^1 = \omega^2 \]

### Step 4: Form the new quadratic equation

Now we have the roots \( \alpha^{2020} = \omega \) and \( \beta^{2020} = \omega^2 \).

The sum of the roots is:

\[ \omega + \omega^2 = -1 \quad \text{(since } 1 + \omega + \omega^2 = 0\text{)} \]

The product of the roots is:

\[ \omega \cdot \omega^2 = \omega^3 = 1 \]

Using the standard form of a quadratic equation \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \):

\[ x^2 - (-1)x + 1 = 0 \]

This simplifies to:

\[ x^2 + x + 1 = 0 \]

### Final Answer

The equation whose roots are \( \alpha^{2020} \) and \( \beta^{2020} \) is:

\[ \boxed{x^2 + x + 1 = 0} \]