Question

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

• A. - 64
• B. 64
• C. - 64
• D. -64

Answer 1

The correct option is B

64

We will express ( 1+ ${\omega }^2-\omega )$ as a single team , because the answers are just one term and are in terms of $\omega$ or ${\omega }^2$ .
1 + $\omega$ + ${\omega }^2$ = 0
1 + ${\omega }^2$ = $-\omega$
1 - ${\omega }^2$ - $\omega$ = - $\omega$ - $\omega$
= -2 $\omega$
${(1+{\omega }^2-\omega )}^6$ = ${(-2\omega )}^6$
= ${(-2)}^6$${\omega }^6$
= $2^6$ ${\left({\omega }^3\right)}^2$
= $2^6$ (because ${\omega }^3$) = 1
= 64