If `x^2-x+1=0` then the value of `sum_[n=1]^[5][x^n+1/x^n]^2` is:

Options

A. 8
B. 10
C. 12
D. None of these

Answer

To solve the equation \( x^2 - x + 1 = 0 \) and find the value of \( \sum_{n=1}^{5} \left( x^n + \frac{1}{x^n} \right)^2 \), we can follow these steps:

### Step 1: Find the roots of the equation
The roots of the quadratic equation \( x^2 - x + 1 = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -1, c = 1 \):
\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2} \]
Thus, the roots are:
\[ x_1 = \frac{1 + i\sqrt{3}}{2}, \quad x_2 = \frac{1 - i\sqrt{3}}{2} \]

### Step 2: Calculate \( x^n + \frac{1}{x^n} \)
Using the property of complex numbers, we can express \( x^n + \frac{1}{x^n} \) in terms of \( x \):
\[ x^n + \frac{1}{x^n} = 2 \cos\left( n \theta \right) \]
where \( \theta \) is the argument of \( x \). For our roots, we can find \( \theta \) as follows:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]
Thus, we have:
\[ x^n + \frac{1}{x^n} = 2 \cos\left( n \frac{\pi}{3} \right) \]

### Step 3: Calculate \( \sum_{n=1}^{5} \left( x^n + \frac{1}{x^n} \right)^2 \)
Now we need to compute:
\[ \sum_{n=1}^{5} \left( 2 \cos\left( n \frac{\pi}{3} \right) \right)^2 = 4 \sum_{n=1}^{5} \cos^2\left( n \frac{\pi}{3} \right) \]
Calculating \( \cos^2\left( n \frac{\pi}{3} \right) \) for \( n = 1, 2, 3, 4, 5 \):
- \( n = 1: \cos^2\left( \frac{\pi}{3} \right) = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \)
- \( n = 2: \cos^2\left( \frac{2\pi}{3} \right) = \left( -\frac{1}{2} \right)^2 = \frac{1}{4} \)
- \( n = 3: \cos^2\left( \frac{3\pi}{3} \right) = \cos^2(\pi) = 1 \)
- \( n = 4: \cos^2\left( \frac{4\pi}{3} \right) = \left( -\frac{1}{2} \right)^2 = \frac{1}{4} \)
- \( n = 5: \cos^2\left( \frac{5\pi}{3} \right) = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \)

Now, summing these values:
\[ \sum_{n=1}^{5} \cos^2\left( n \frac{\pi}{3} \right) = \frac{1}{4} + \frac{1}{4} + 1 + \frac{1}{4} + \frac{1}{4} = \frac{5}{4} + 1 = \frac{9}{4} \]

### Step 4: Final calculation
Now substituting back:
\[ 4 \sum_{n=1}^{5} \cos^2\left( n \frac{\pi}{3} \right) = 4 \cdot \frac{9}{4} = 9 \]

### Conclusion
Thus, the value of \( \sum_{n=1}^{5} \left( x^n + \frac{1}{x^n} \right)^2 \) is:
\[ \boxed{9} \]