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