Question

If $\omega$ is a complex cube root of unity, then $\left(1-\omega+\omega^{2}\right)^{6}+\left(1-\omega^{2}+\omega\right)^{6}=$

• A. 0
• B. 6
• C. 64
• D. 128

Answer 1

Answer: (D)

We have, $\left(1-\omega+\omega^{2}\right)^{6}+\left(1-\omega^{2}+\omega\right)^{6}$
$=(-\omega-\omega)^{6}+\left(-\omega^{2}-\omega^{2}\right)^{6} $
$\left[\because 1+\omega+\omega^{2}=0\right]$
$=(-2 \omega)^{6}+\left(-2 \omega^{2}\right)^{6}$
$=64 \omega^{6}+64 \omega^{12}$
$=64\left(\omega^{3}\right)^{2}+64\left(\omega^{3}\right)^{4}$
$=64(1)^{2}+64(1)^{4}$
$=64+64=128$