Question

If $\omega$ is a cube root of unity, then the value of $(1\omega +{\omega }^2{)}^5+(1+\omega {\omega }^2{)}^5$=

• A. $16$
• B. $32$
• C. $48$
• D. $64$

Answer 1

Answer: (B)

$32$

${(1-w+w^2)}^5+{(1+w+w^2)}^5$ $1+w+w^2=0$

$\Rightarrow {(1-(-1))}^5=0$ $\therefore w+w^2=-1$

$=2^5$

$=32.$

Hence, the answer is $32.$