If `omega` is a complex non-real cube root of unity, then `omega` satisfies which one of the following equations ?

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`

Explanation

To solve the problem, we need to determine which equation is satisfied by the complex non-real cube root of unity, denoted as \( \omega \).

### Step-by-Step Solution:

1. **Understanding Cube Roots of Unity**:
The cube roots of unity are the solutions to the equation \( x^3 = 1 \). These roots can be expressed as:
\[ 1, \omega, \omega^2 \]
where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{4\pi i / 3} \). The value \( \omega \) is a non-real cube root of unity.

2. **Properties of Cube Roots of Unity**:
The sum of the cube roots of unity is:
\[ 1 + \omega + \omega^2 = 0 \]
From this, we can derive that:
\[ \omega + \omega^2 = -1 \]

3. **Finding the Polynomial**:
Since \( \omega \) is a root of the polynomial formed by the cube roots of unity, we can express this as:
\[ x^3 - 1 = 0 \]
Factoring this gives:
\[ (x - 1)(x^2 + x + 1) = 0 \]
Thus, the non-real cube roots of unity \( \omega \) and \( \omega^2 \) are the roots of the quadratic polynomial:
\[ x^2 + x + 1 = 0 \]

4. **Verifying the Equation**:
Since \( \omega \) is a root of \( x^2 + x + 1 = 0 \), we can conclude that:
\[ \omega^2 + \omega + 1 = 0 \]

5. **Conclusion**:
Therefore, the equation that \( \omega \) satisfies is:
\[ x^2 + x + 1 = 0 \]

### Final Answer:
The equation satisfied by \( \omega \) is:
\[ x^2 + x + 1 = 0 \]