If `omega` is a complex cube root of unity, then what is `omega^(10)+omega^(-10)` equal to ?

Options

A. 2
B. `-1`
C. `-2`
D. 1

Explanation

To solve the problem, we need to find the value of \( \omega^{10} + \omega^{-10} \), where \( \omega \) is a complex cube root of unity. The cube roots of unity are the solutions to the equation \( x^3 = 1 \), which are \( 1, \omega, \) and \( \omega^2 \), where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{4\pi i / 3} \).

### Step-by-step Solution:

1. **Understanding the properties of \( \omega \)**:
- Since \( \omega \) is a cube root of unity, we have:
\[ \omega^3 = 1 \]
- This implies that \( \omega^n \) can be simplified based on the remainder when \( n \) is divided by 3.

2. **Calculate \( \omega^{10} \)**:
- To find \( \omega^{10} \), we divide 10 by 3:
\[ 10 \div 3 = 3 \quad \text{(remainder 1)} \]
- Thus, \( \omega^{10} = \omega^{3 \cdot 3 + 1} = (\omega^3)^3 \cdot \omega^1 = 1^3 \cdot \omega = \omega \).

3. **Calculate \( \omega^{-10} \)**:
- To find \( \omega^{-10} \), we can use the property of exponents:
\[ \omega^{-10} = \frac{1}{\omega^{10}} = \frac{1}{\omega} \]
- Since \( \omega^3 = 1 \), we also know that \( \omega^2 = \frac{1}{\omega} \).

4. **Combine the results**:
- Now we can find \( \omega^{10} + \omega^{-10} \):
\[ \omega^{10} + \omega^{-10} = \omega + \frac{1}{\omega} = \omega + \omega^2 \]

5. **Use the property of cube roots of unity**:
- From the properties of the cube roots of unity, we know:
\[ 1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1 \]

6. **Final result**:
- Therefore, we conclude that:
\[ \omega^{10} + \omega^{-10} = -1 \]

### Summary:
The value of \( \omega^{10} + \omega^{-10} \) is \( -1 \).