If `alpha` is a root of `x^(2)-x+1=0` and `beta` be the digit at unit place in the number `2^(2^(n))-5,n in N(n gt 1)`, then `alpha^(2005)+alpha^(-2005)+beta^(2006)+beta^(-2006)` is equal to ____________

Answer

To solve the problem, we need to evaluate the expression:

\[ \alpha^{2005} + \alpha^{-2005} + \beta^{2006} + \beta^{-2006} \]

where \(\alpha\) is a root of the equation \(x^2 - x + 1 = 0\) and \(\beta\) is the unit digit of \(2^{2^n} - 5\) for \(n \in \mathbb{N}\) and \(n > 1\).

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

We start with the quadratic equation:

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

Using the quadratic formula, the roots are given by:

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

Here, \(a = 1\), \(b = -1\), and \(c = 1\). Substituting these values:

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

This simplifies to:

\[ x = \frac{1 \pm i\sqrt{3}}{2} \]

Thus, the roots are:

\[ \alpha_1 = \frac{1 + i\sqrt{3}}{2}, \quad \alpha_2 = \frac{1 - i\sqrt{3}}{2} \]

### Step 2: Calculate \(\alpha^{2005} + \alpha^{-2005}\)

Using the exponential form, we can express \(\alpha\) in terms of Euler's formula:

\[ \alpha = e^{i\pi/3} \quad \text{and} \quad \alpha^{-1} = e^{-i\pi/3} \]

Thus, we can write:

\[ \alpha^{2005} = \left(e^{i\pi/3}\right)^{2005} = e^{i \cdot \frac{2005\pi}{3}} \quad \text{and} \quad \alpha^{-2005} = \left(e^{-i\pi/3}\right)^{2005} = e^{-i \cdot \frac{2005\pi}{3}} \]

Now, we need to simplify \(e^{i \cdot \frac{2005\pi}{3}}\):

Calculating \(2005 \mod 6\) (since \(2\pi\) has a period of \(6\)):

\[ 2005 \div 6 = 334 \quad \text{remainder } 1 \]

Thus,

\[ \frac{2005\pi}{3} = 334 \cdot 2\pi + \frac{\pi}{3} \]

So,

\[ e^{i \cdot \frac{2005\pi}{3}} = e^{i \cdot \frac{\pi}{3}} \quad \text{and} \quad e^{-i \cdot \frac{2005\pi}{3}} = e^{-i \cdot \frac{\pi}{3}} \]

Now we can combine these:

\[ \alpha^{2005} + \alpha^{-2005} = e^{i\frac{\pi}{3}} + e^{-i\frac{\pi}{3}} = 2\cos\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1 \]

### Step 3: Find \(\beta\)

Next, we need to find \(\beta\), the unit digit of \(2^{2^n} - 5\) for \(n > 1\).

Calculating \(2^{2^n}\) for \(n = 2, 3, 4\):

- For \(n = 2\): \(2^{2^2} = 2^4 = 16\) (unit digit is 6)
- For \(n = 3\): \(2^{2^3} = 2^8 = 256\) (unit digit is 6)
- For \(n = 4\): \(2^{2^4} = 2^{16} = 65536\) (unit digit is 6)

Thus, the unit digit of \(2^{2^n}\) is always 6 for \(n > 1\). Therefore, the unit digit of \(2^{2^n} - 5\) is:

\[ 6 - 5 = 1 \]

So, \(\beta = 1\).

### Step 4: Calculate \(\beta^{2006} + \beta^{-2006}\)

Since \(\beta = 1\):

\[ \beta^{2006} = 1^{2006} = 1 \quad \text{and} \quad \beta^{-2006} = 1^{-2006} = 1 \]

Thus,

\[ \beta^{2006} + \beta^{-2006} = 1 + 1 = 2 \]

### Final Step: Combine results

Finally, we combine the results:

\[ \alpha^{2005} + \alpha^{-2005} + \beta^{2006} + \beta^{-2006} = 1 + 2 = 3 \]

Thus, the final answer is:

\[ \boxed{3} \]