Question

If \( \omega \) is the cube root of unity, then: \[ \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} \]

• A. \( 2 \)
• B. \( -2 \)
• C. \( 1 \)
• D. \( -1 \)

Hint:

For cube roots of unity, use the relation \( \omega^2 + \omega + 1 = 0 \) to simplify expressions involving \( \omega \) and \( \omega^2 \).

Answer 1

The Correct Option is D

We are given that \( \omega \) is the cube root of unity. The cube roots of unity satisfy the following relations: \[ \omega^3 = 1, \quad \omega^2 + \omega + 1 = 0. \] This implies that: \[ \omega^2 = -\omega - 1. \] 
Step 1: We need to evaluate the expression: \[ \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}. \] Using the fact that \( \omega^2 = -\omega - 1 \), we can substitute this in the numerator and denominator: \[ {Numerator:} \quad a + b\omega + c\omega^2 = a + b\omega + c(-\omega - 1) = a + b\omega - c\omega - c = a - c + (b - c)\omega, \] \[ {Denominator:} \quad c + a\omega + b\omega^2 = c + a\omega + b(-\omega - 1) = c + a\omega - b\omega - b = (c - b) + (a - b)\omega. \] Thus, the expression becomes: \[ \frac{a - c + (b - c)\omega}{(c - b) + (a - b)\omega}. \] 
Step 2: Now, we evaluate the second part of the equation: \[ \frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2}. \] Similarly, applying \( \omega^2 = -\omega - 1 \), we get: \[ {Numerator:} \quad a + b\omega + c\omega^2 = a + b\omega + c(-\omega - 1) = a - c + (b - c)\omega, \] \[ {Denominator:} \quad b + c\omega + a\omega^2 = b + c\omega + a(-\omega - 1) = b - a + (c - a)\omega. \] Thus, this expression simplifies to: \[ \frac{a - c + (b - c)\omega}{b - a + (c - a)\omega}. \] 
Step 3: Since both expressions are now in the same form, it follows that: \[ \frac{a - c + (b - c)\omega}{(c - b) + (a - b)\omega} = \frac{a - c + (b - c)\omega}{b - a + (c - a)\omega}. \] This implies that the value of the expression is \( -1 \).