If ω ≠ 1 is a cube root of unity, then what is (1+ω-ω^2 )^100+ (1-ω+ω^2 )^100 equal to?

Correct answer: (D) -2^100. Let I= (1+ω-ω^2 )^100+ (1-ω+ω^2 )^100 = (-ω^2-ω^2 )^100+(-ω-ω)^100 \∵ 1+ω+ω^2=0 \ =2^100 (ω^200+ω^100 ) =2^100 (ω^2+ω ) \∵ ω^3=1 \ =-2^100 .

Q4 / 120 · NDA 2024 (Phase 2) · Mathematics

If $\omega \neq 1$ is a cube root of unity, then what is $\left(1+\omega-\omega^2\right)^{100}+\left(1-\omega+\omega^2\right)^{100}$ equal to?

A
$2^{100} \omega^2$
B
$2^{100} \omega$
C
$2^{100}$
Correct answer
D
$-2^{100}$

Solution

$\begin{aligned} & \text { Let } I=\left(1+\omega-\omega^2\right)^{100}+\left(1-\omega+\omega^2\right)^{100} \\ & =\left(-\omega^2-\omega^2\right)^{100}+(-\omega-\omega)^{100}\left\{\because 1+\omega+\omega^2=0\right\} \\ & =2^{100}\left(\omega^{200}+\omega^{100}\right) \\ & =2^{100}\left(\omega^2+\omega\right)\left\{\because \omega^3=1\right\} \\ & =-2^{100} .\end{aligned}$

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

If $\omega \neq 1$ is a cube root of unity, then what is $\left(1+\omega-\omega^2\right)^{100}+\left(1-\omega+\omega^2\right)^{100}$ equal to?

Solution

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