If ` omega^(3) = 1 and omega ne 1 ` then `(1+omega)(1+omega^2)(1+omega^4)(1+omega^5)` is equal to

Options

A. 3
B. `-3`
C. 9
D. 1

Answer

To solve the problem, we need to evaluate the expression \((1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^5)\) given that \(\omega^3 = 1\) and \(\omega \neq 1\).

### Step-by-Step Solution:

1. **Understanding the Roots of Unity**:
Since \(\omega^3 = 1\), \(\omega\) is a primitive cube root of unity. The cube roots of unity are \(1\), \(\omega\), and \(\omega^2\). Here, \(\omega\) and \(\omega^2\) are the non-real roots.

2. **Identifying the Powers**:
Notice that \(\omega^4\) and \(\omega^5\) can be simplified using the property of cube roots:
\[ \omega^4 = \omega^{3+1} = \omega^3 \cdot \omega = 1 \cdot \omega = \omega \]
\[ \omega^5 = \omega^{3+2} = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2 \]

3. **Substituting Back**:
Now we can rewrite the expression:
\[ (1+\omega)(1+\omega^2)(1+\omega)(1+\omega^2) \]
This simplifies to:
\[ (1+\omega)^2(1+\omega^2)^2 \]

4. **Calculating Each Factor**:
We can calculate \(1+\omega\) and \(1+\omega^2\):
- For \(1 + \omega\):
\[ 1 + \omega = -\omega^2 \quad (\text{since } 1 + \omega + \omega^2 = 0) \]
- For \(1 + \omega^2\):
\[ 1 + \omega^2 = -\omega \quad (\text{since } 1 + \omega + \omega^2 = 0) \]

5. **Substituting Values**:
Now substituting these back into the expression:
\[ (1+\omega)^2 = (-\omega^2)^2 = \omega^4 = 1 \]
\[ (1+\omega^2)^2 = (-\omega)^2 = \omega^2 \]

6. **Final Calculation**:
Therefore, we have:
\[ (1+\omega)^2(1+\omega^2)^2 = 1 \cdot \omega^2 = \omega^2 \]

7. **Conclusion**:
Since \(\omega^3 = 1\), we know that \(\omega^2\) is simply \(\omega^2\). Thus, the final answer is:
\[ \boxed{1} \]