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If $\mathrm{\alpha, \beta, \gamma, \delta}$ are the roots of the equation $\mathrm{x^{4}+x^{3}+x^{2}+x+1=0}$ , then $\mathrm{\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}}$ is equal to :
Option: 1
$-4$

Option: 2
$-1$

Option: 3
$1$

Option: 4
$4$

Answers (1)

best_answer

$\mathrm{\text {Let } x^{4}+x^{3}+x^{2}+x-1=0 }\\$

$\mathrm{\frac{x^{5}-1}{x-1}=0 \quad(x \neq 1) }\\$

$\mathrm{x^{5}-1=0} \\$

$\mathrm{x=(1)^{1 / 5}} \\$

$\mathrm{x=(\cos 2 r \pi+i \sin 2 r \pi)^{1 / 5}} \\$

$\mathrm{x=\cos \frac{2 r \pi}{5}+i \sin \frac{2 r \pi}{5}}$

Putting $\mathrm{r=0,1,2,3,4}$ the five roots are given by

$\mathrm{\cos 0+i \sin 0=1, \quad \alpha=\cos \frac{2 \pi}{5}+i \sin \frac{2 \pi}{5}}\\$

$\mathrm{\beta=\cos \frac{4 \pi}{5}+i \sin \frac{4 \pi}{5}, \gamma=\cos \frac{6 \pi}{5}+i \sin \frac{6 \pi}{5}}\\$

$\mathrm{\delta=\cos \frac{\theta \pi}{5}+i \sin \frac{8 \pi}{5}}\\$

$\mathrm{\text { then } \alpha^{2021}+\beta^{2021}+y^{2021}+$ $\delta^{2021}=\alpha+\beta+\gamma +\delta\\$

$=-1$

using De-morves's theroem

$\mathrm{(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta}$

hence correct option is 2

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Sayak

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