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(B) The given equation is $x^{4}+x^{3}+x^{2}+x+1=0$.This is a geometric series sum,which can be written as $\frac{x^{5}-1}{x-1} = 0$ for $x 
eq 1$.Th

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$-1$

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(B) The given equation is $x^{4}+x^{3}+x^{2}+x+1=0$.This is a geometric series sum,which can be written as $\frac{x^{5}-1}{x-1} = 0$ for $x 
eq 1$.The roots $\alpha, \beta, \gamma, \delta$ are the $5^{	ext{th}}$ roots of unity excluding $1$,i.e.,$\omega, \omega^{2}, \omega^{3}, \omega^{4}$ where $\omega = e^{i \frac{2\pi}{5}}$.Since $\omega^{5} = 1$,we have $\omega^{2021} = (\omega^{5})^{404} \cdot \omega = \omega$.Similarly,$\beta^{2021} = \omega^{2}$,$\gamma^{2021} = \omega^{3}$,and $\delta^{2021} = \omega^{4}$.Thus,the sum is $\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021} = \omega + \omega^{2} + \omega^{3} + \omega^{4}$.From the equation $x^{4}+x^{3}+x^{2}+x+1=0$,the sum of the roots is $\alpha+\beta+\gamma+\delta = -\frac{1}{1} = -1$.

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

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