If `p^(2)-p+1=0`, then the value of `p^(3n)` can be

Options

A. 1
B. `-1`
C. 0
D. `cos npi`

Answer

To solve the equation \( p^2 - p + 1 = 0 \) and find the value of \( p^{3n} \), we can follow these steps:

### Step 1: Find the roots of the quadratic equation
We start with the quadratic equation:
\[ p^2 - p + 1 = 0 \]
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = 1 \):
\[ p = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ p = \frac{1 \pm \sqrt{1 - 4}}{2} \]
\[ p = \frac{1 \pm \sqrt{-3}}{2} \]
\[ p = \frac{1 \pm i\sqrt{3}}{2} \]

### Step 2: Express the roots in polar form
The two roots can be expressed as:
\[ p_1 = \frac{1}{2} + i\frac{\sqrt{3}}{2} \quad \text{and} \quad p_2 = \frac{1}{2} - i\frac{\sqrt{3}}{2} \]
These can be represented in polar form as:
\[ p_1 = e^{i\frac{\pi}{3}} \quad \text{and} \quad p_2 = e^{-i\frac{\pi}{3}} \]

### Step 3: Calculate \( p^{3n} \)
Now, we calculate \( p^{3n} \):
For \( p_1 \):
\[ p_1^{3n} = \left(e^{i\frac{\pi}{3}}\right)^{3n} = e^{i\pi n} \]
For \( p_2 \):
\[ p_2^{3n} = \left(e^{-i\frac{\pi}{3}}\right)^{3n} = e^{-i\pi n} \]

### Step 4: Simplify \( e^{i\pi n} \) and \( e^{-i\pi n} \)
Using Euler's formula:
\[ e^{i\pi n} = \cos(\pi n) + i\sin(\pi n) \]
\[ e^{-i\pi n} = \cos(-\pi n) + i\sin(-\pi n) = \cos(\pi n) - i\sin(\pi n) \]
Both expressions yield:
\[ p^{3n} = \cos(\pi n) \quad \text{(for both roots)} \]

### Step 5: Determine the values based on \( n \)
- If \( n \) is even, \( \cos(\pi n) = 1 \).
- If \( n \) is odd, \( \cos(\pi n) = -1 \).

### Conclusion
Thus, the value of \( p^{3n} \) can be:
- \( 1 \) when \( n \) is even
- \( -1 \) when \( n \) is odd