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

Correct answer: (D) 32. (1+ω-ω^2 )^5+ (1-ω+ω^2 )^5 = (-ω^2-ω^2 )^5+(-ω-ω)^5 =-32 (ω^2 )^5+(-32) ω^5 =-32 (ω+ω^2 )=32

Q97 / 170 · Manipal (MET) 2017 · Mathematics

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

A
16
B
8
C
9
Correct answer
D
32

Solution

$\begin{aligned} &\left(1+\omega-\omega^2\right)^5+\left(1-\omega+\omega^2\right)^5 \\ &=\left(-\omega^2-\omega^2\right)^5+(-\omega-\omega)^5 \\ &=-32\left(\omega^2\right)^5+(-32) \omega^5 \\ &=-32\left(\omega+\omega^2\right)=32\end{aligned}$

Hence, the correct answer is option: 'D'.

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

Solution

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