If `alpha, beta` are the roots of `x^(2) - x + 1 = 0` then the quadratic equation whose roots are `alpha^(2015), beta^(2015)` 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 quadratic equation whose roots are \( \alpha^{2015} \) and \( \beta^{2015} \), given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - x + 1 = 0 \).

### Step-by-step Solution:

1. **Identify the roots of the given quadratic equation**:
The quadratic equation is \( x^2 - x + 1 = 0 \).
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -1, c = 1 \).

Substituting these values into the formula:
\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2} \]

Thus, the roots are:
\[ \alpha = \frac{1 + i\sqrt{3}}{2}, \quad \beta = \frac{1 - i\sqrt{3}}{2} \]

2. **Express the roots in terms of cube roots of unity**:
The roots \( \alpha \) and \( \beta \) can be expressed as:
\[ \alpha = -\omega, \quad \beta = -\omega^2 \]
where \( \omega = e^{2\pi i / 3} \) is a cube root of unity.

3. **Calculate \( \alpha^{2015} \) and \( \beta^{2015} \)**:
We need to find \( \alpha^{2015} \) and \( \beta^{2015} \):
\[ \alpha^{2015} = (-\omega)^{2015} = -\omega^{2015} \]
Since \( \omega^3 = 1 \), we can reduce the exponent modulo 3:
\[ 2015 \mod 3 = 1 \quad \text{(since } 2015 = 671 \times 3 + 2\text{)} \]
Thus,
\[ \alpha^{2015} = -\omega^2 \]

Similarly for \( \beta \):
\[ \beta^{2015} = (-\omega^2)^{2015} = -(\omega^2)^{2015} = -\omega^{4030} \]
Reducing \( 4030 \mod 3 \):
\[ 4030 \mod 3 = 1 \]
Thus,
\[ \beta^{2015} = -\omega \]

4. **Find the new roots**:
The new roots are:
\[ \alpha^{2015} = -\omega^2, \quad \beta^{2015} = -\omega \]

5. **Form the new quadratic equation**:
The sum of the roots \( S \) and product of the roots \( P \) are:
\[ S = -\omega^2 - \omega = -(\omega + \omega^2) = -(-1) = 1 \]
\[ P = (-\omega^2)(-\omega) = \omega^3 = 1 \]

Therefore, the quadratic equation with roots \( \alpha^{2015} \) and \( \beta^{2015} \) is:
\[ x^2 - Sx + P = x^2 - 1x + 1 = x^2 - x + 1 = 0 \]

### Final Answer:
The quadratic equation whose roots are \( \alpha^{2015} \) and \( \beta^{2015} \) is:
\[ x^2 - x + 1 = 0 \]