If `1,alpha_1,alpha_2,alpha_3,.........,alpha_(3n)` be the roots of the eqution `x^(3n+1) - 1 =0`, and w be an imaginary cube root of unity, then `((w^2-alpha_1)(w^2-alpha_2)....(w^(3n)-alpha_(3n))) /((w-alpha_1)(w2-alpha)....(w-alpha_(3n)))`

Options

A. `omega`
B. `-omega`
C. 1
D. `omega^(2)`

Answer

To solve the problem, we need to simplify the expression given in the question. Let's break it down step by step.

### Step 1: Understand the Roots
The equation given is \( x^{3n+1} - 1 = 0 \). The roots of this equation are the \( (3n+1) \)-th roots of unity, which are given as \( 1, \alpha_1, \alpha_2, \ldots, \alpha_{3n} \).

### Step 2: Write the Polynomial in Terms of its Roots
We can express the polynomial as:
\[ x^{3n+1} - 1 = (x - 1)(x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_{3n}) \]

### Step 3: Evaluate the Expression
We need to simplify the expression:
\[ \frac{(w^2 - \alpha_1)(w^2 - \alpha_2) \cdots (w^2 - \alpha_{3n})}{(w - \alpha_1)(w - \alpha_2) \cdots (w - \alpha_{3n})} \]
where \( w \) is an imaginary cube root of unity.

### Step 4: Use the Polynomial Form
Let’s denote:
\[ P(x) = x^{3n+1} - 1 \]
Then, we can evaluate \( P(w^2) \) and \( P(w) \):
\[ P(w^2) = w^{2(3n+1)} - 1 = (w^3)^{2n + \frac{2}{3}} - 1 = 1 - 1 = 0 \]
\[ P(w) = w^{3n+1} - 1 = 1 - 1 = 0 \]

### Step 5: Substitute into the Polynomial
Now, substituting \( x = w^2 \) and \( x = w \) into the polynomial gives us:
\[ P(w^2) = (w^2 - 1)(w^2 - \alpha_1)(w^2 - \alpha_2) \cdots (w^2 - \alpha_{3n}) \]
\[ P(w) = (w - 1)(w - \alpha_1)(w - \alpha_2) \cdots (w - \alpha_{3n}) \]

### Step 6: Form the Ratio
Taking the ratio of these two evaluations gives:
\[ \frac{P(w^2)}{P(w)} = \frac{(w^2 - 1)(w^2 - \alpha_1)(w^2 - \alpha_2) \cdots (w^2 - \alpha_{3n})}{(w - 1)(w - \alpha_1)(w - \alpha_2) \cdots (w - \alpha_{3n})} \]

### Step 7: Simplify
Using the identity \( a^2 - b^2 = (a - b)(a + b) \), we can simplify the expression:
\[ = \frac{(w^2 - 1)}{(w - 1)} \cdot \frac{(w^2 - \alpha_1)(w^2 - \alpha_2) \cdots (w^2 - \alpha_{3n})}{(w - \alpha_1)(w - \alpha_2) \cdots (w - \alpha_{3n})} \]

### Step 8: Final Evaluation
Since \( w^2 - 1 = (w - 1)(w + 1) \), the \( (w - 1) \) terms cancel out:
\[ = (w + 1) \cdot \frac{(w^2 - \alpha_1)(w^2 - \alpha_2) \cdots (w^2 - \alpha_{3n})}{(w - \alpha_1)(w - \alpha_2) \cdots (w - \alpha_{3n})} \]

### Conclusion
After evaluating the above expression, we find that the final result simplifies to \( 1 \).