The value of `(x+omega+omega^(2)) (x +omega^(2)+omega^(4))(x +omega^(4)+omega^(8))`.... till 2n factors

Options

A. `(x-1)^(2n)`
B. `(x-1)^(2n+1)`
C. `(x-1)^(2n-1)`
D. `(X-1)^(2n+2)`

Answer

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

\[ (x + \omega + \omega^2)(x + \omega^2 + \omega^4)(x + \omega^4 + \omega^8) \ldots \text{ (up to 2n factors)} \]

where \(\omega\) is a primitive cube root of unity. This means that \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).

### Step-by-Step Solution:

1. **Understanding the Roots of Unity:**
Since \(\omega\) is a cube root of unity, we have:
\[ \omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0 \]
This implies:
\[ \omega + \omega^2 = -1 \]

2. **Simplifying Each Factor:**
Let's simplify each factor in the product:
- The first factor is:
\[ x + \omega + \omega^2 = x - 1 \]
- The second factor is:
\[ x + \omega^2 + \omega^4 = x + \omega^2 + \omega = x - 1 \]
- The third factor is:
\[ x + \omega^4 + \omega^8 = x + \omega + \omega^2 = x - 1 \]
- Continuing this process, we see that every factor results in \(x - 1\).

3. **Counting the Factors:**
Since we have \(2n\) factors, we can express the product as:
\[ P = (x - 1)(x - 1)(x - 1) \ldots \text{ (2n times)} \]

4. **Final Expression:**
Therefore, the product can be written as:
\[ P = (x - 1)^{2n} \]

### Conclusion:
The value of the expression \((x + \omega + \omega^2)(x + \omega^2 + \omega^4)(x + \omega^4 + \omega^8) \ldots\) up to \(2n\) factors is:
\[ \boxed{(x - 1)^{2n}} \]