Question

If \( \omega \) is the complex cube root of unity and \[ \left( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} \right)^k + \left( \frac{a + b\omega + c\omega^2}{b + a\omega^2 + c\omega} \right)^2 = 2, \] then \( 2k + 1 \) is always:

• A. divisible by 2
• B. divisible by 6
• C. divisible by 3
• D. divisible by 5
  \

Hint:

For problems involving cube roots of unity, use the identity \( 1 + \omega + \omega^2 = 0 \) to simplify expressions and find cyclic properties of exponents.

Answer 1

The Correct Option is C

We are given the equation: \[ \left( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} \right)^k + \left( \frac{a + b\omega + c\omega^2}{b + a\omega^2 + c\omega} \right)^2 = 2. \] where \( \omega \) is the complex cube root of unity, satisfying: \[ \omega^3 = 1, \quad 1 + \omega + \omega^2 = 0. \] Step 1: Consider the Expression Properties
Since \( \omega \) is a cube root of unity, applying its properties simplifies the fractions into cyclic expressions. From algebraic manipulation, the given equation simplifies to: \[ x^k + x^2 = 2. \] Since \( x \) is a cube root of unity, we solve for possible values of \( k \). Step 2: Identify Divisibility of \( 2k+1 \)
Through solving and checking integer values, we find that: \[ 2k + 1 \text{ is always divisible by } 3. \] Step 3: Conclusion
Thus, we conclude: \[ \boxed{2k+1 \text{ is divisible by 3}.} \] \bigskip