If `1,alpha,alpha^(2),……….,alpha^(n-1)` are `n^(th)` root of unity, the value of `(3-alpha)(3-alpha^(2))(3-alpha^(3))……(3-alpha^(n-1))`, is

Options

A. n
B. 0
C. `(3n-1)/2`
D. `(3n+1)/2`

Explanation

To solve the problem, we need to find the value of the expression \((3 - \alpha)(3 - \alpha^2)(3 - \alpha^3) \ldots (3 - \alpha^{n-1})\), where \(1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\) are the \(n^{th}\) roots of unity.

### Step-by-Step Solution:

1. **Understanding the Roots of Unity**:
The \(n^{th}\) roots of unity are given by the formula:
\[ \alpha_k = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
Here, \(1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\) correspond to these roots.

2. **Using the Polynomial Representation**:
The \(n^{th}\) roots of unity can be represented by the polynomial:
\[ P(x) = x^n - 1 = (x - 1)(x - \alpha)(x - \alpha^2) \ldots (x - \alpha^{n-1}) \]

3. **Finding the Expression**:
We want to evaluate:
\[ (3 - \alpha)(3 - \alpha^2)(3 - \alpha^3) \ldots (3 - \alpha^{n-1}) \]
This can be expressed as:
\[ P(3) = 3^n - 1 \]

4. **Relating the Expression to the Polynomial**:
We can rewrite our expression as:
\[ P(3) = (3 - 1)(3 - \alpha)(3 - \alpha^2) \ldots (3 - \alpha^{n-1}) = 2 \cdot (3 - \alpha)(3 - \alpha^2) \ldots (3 - \alpha^{n-1}) \]

5. **Solving for the Desired Product**:
Thus, we have:
\[ (3 - \alpha)(3 - \alpha^2) \ldots (3 - \alpha^{n-1}) = \frac{P(3)}{2} \]
Substituting \(P(3)\):
\[ (3 - \alpha)(3 - \alpha^2) \ldots (3 - \alpha^{n-1}) = \frac{3^n - 1}{2} \]

### Final Answer:
The value of \((3 - \alpha)(3 - \alpha^2)(3 - \alpha^3) \ldots (3 - \alpha^{n-1})\) is:
\[ \frac{3^n - 1}{2} \]