Question

If \(1, \alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_{n-1}\) are n roots of the equation, \( x^n = 1 \) then the value of \( (1-\alpha_1)(1-\alpha_2)(1-\alpha_3)\ldots(1-\alpha_{n-1}) \) is

• A. n
• B. n-1
• C. n-2
• D. n-3

Hint:

This is a standard result related to roots of unity. The polynomial \( \Phi_n(x) = \frac{x^n-1}{x-1} \) is called the nth cyclotomic polynomial, and its roots are the primitive nth roots of unity (for prime n). The value of this polynomial at \(x=1\) gives the desired product, which is always \(n\).

Answer 1

The Correct Option is A

Step 1: Concept Definition:
The equation \(x^n = 1\) defines the \(n^{th}\) roots of unity. The problem requires calculating the product of terms of the form \( (1 - \alpha_k) \), where \( \alpha_k \) represents the roots of unity excluding 1.

Step 2: Core Principle:
The polynomial \( P(x) = x^n - 1 \) has the \(n\) roots of unity as its roots. This polynomial can be factored as:
\[ x^n - 1 = (x-1)(x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_{n-1}) \]
Alternatively, using the geometric series sum formula:
\[ x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \ldots + x + 1) \]
Comparing these factorizations yields an expression for the product involving the roots.

Step 3: Detailed Derivation:
Equating the non-\( (x-1) \) components of the two factorizations for \( x^n - 1 \):
\[ (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_{n-1}) = x^{n-1} + x^{n-2} + \ldots + x + 1 \]
The objective is to find the value of \( (1-\alpha_1)(1-\alpha_2)\ldots(1-\alpha_{n-1}) \).
This is achieved by substituting \( x = 1 \) into the aforementioned polynomial identity.
Let \( Q(x) = (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_{n-1}) \).
We need to compute \( Q(1) \).
\[ Q(1) = 1^{n-1} + 1^{n-2} + \ldots + 1^1 + 1 \]
The right-hand side summation consists of \(n\) terms, each equaling 1.
\[ Q(1) = \underbrace{1 + 1 + \ldots + 1}_{n \text{ terms}} = n \]

Step 4: Conclusion:
The resultant value of the expression is \( n \).