If `alpha, beta` are the roots of `x^(2)+x+1=0`, then the equation whose roots are `alpha^(5), beta^(5)` is

Options

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

Answer

To find the equation whose roots are \( \alpha^5 \) and \( \beta^5 \), given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + x + 1 = 0 \), we can follow these steps:

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

The roots of the equation \( x^2 + x + 1 = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = 1 \).

Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, the roots are complex:
\[ x = \frac{-1 \pm \sqrt{-3}}{2 \cdot 1} = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, we have:
\[ \alpha = \frac{-1 + i\sqrt{3}}{2}, \quad \beta = \frac{-1 - i\sqrt{3}}{2} \]

### Step 2: Calculate \( \alpha^5 \) and \( \beta^5 \)

To find \( \alpha^5 \) and \( \beta^5 \), we can use the fact that \( \alpha \) and \( \beta \) are cube roots of unity. Specifically, we can express them in terms of \( \omega \) (where \( \omega = e^{2\pi i / 3} \)):
\[ \alpha = \omega, \quad \beta = \omega^2 \]
where \( \omega^3 = 1 \).

Calculating \( \alpha^5 \) and \( \beta^5 \):
\[ \alpha^5 = \omega^5 = \omega^{3} \cdot \omega^{2} = 1 \cdot \omega^{2} = \omega^{2} \]
\[ \beta^5 = (\omega^2)^5 = \omega^{10} = \omega^{9} \cdot \omega = 1 \cdot \omega = \omega \]

### Step 3: Find the new equation with roots \( \alpha^5 \) and \( \beta^5 \)

Now, we need to find the equation whose roots are \( \alpha^5 = \omega^2 \) and \( \beta^5 = \omega \). The sum and product of the roots are:
\[ \text{Sum} = \omega + \omega^2 \]
\[ \text{Product} = \omega \cdot \omega^2 = \omega^3 = 1 \]

Using Vieta's formulas, the new quadratic equation can be formed as:
\[ x^2 - (\text{Sum})x + \text{Product} = 0 \]
Substituting the values:
\[ x^2 - (-1)x + 1 = 0 \quad \text{(since } \omega + \omega^2 = -1\text{)} \]
This simplifies to:
\[ x^2 + x + 1 = 0 \]

### Final Answer

The equation whose roots are \( \alpha^5 \) and \( \beta^5 \) is:
\[ \boxed{x^2 + x + 1 = 0} \]