Question

If $\omega$ is an imaginary cube root of 1, then $(1+\omega-\omega^2)^5 + (1-\omega+\omega^2)^5$ is equal to

• A. $16$
• B. $8$
• C. $9$
• D. $32$

Hint:

$\omega^3=1$, $1+\omega+\omega^2=0$.

Answer 1

The Correct Option is D

Step 1: $1+\omega+\omega^2=0 \Rightarrow 1+\omega = -\omega^2$, $1+\omega^2 = -\omega$.}
Step 2: $1+\omega-\omega^2 = (1+\omega) - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2$. $1-\omega+\omega^2 = (1+\omega^2) - \omega = -\omega - \omega = -2\omega$.}
Step 3: $(-2\omega^2)^5 + (-2\omega)^5 = -32\omega^{10} - 32\omega^5 = -32\omega - 32\omega^2 = -32(\omega+\omega^2)= -32(-1)=32$.}