α^2021 + β^2021 + γ^2021 + δ^2021 for x^4 + x^3 + x^2 + x + 1 = 0

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

(A) –4
(B) –1
(C) 1
(D) 4

Solution:

First, simplify the given equation.

Use property of nth root of unity i.e.

Step 1 of 2:

Given, ${x^4} + {x^3} + {x^2} + x + 1 = 0$ ….(1)

Multiply by x on both sides

${x^5} + {x^4} + {x^3} + {x^2} + x = 0$ …..(2)

From (1), we get ${x^4} + {x^3} + {x^2} + x = - 1$

So, equation (2) becomes

$\begin{gathered} \Rightarrow {x^5} - 1 = 0 \hfill \\ \Rightarrow {x^5} = 1 \hfill \\ \end{gathered}$

Therefore roots are

$1,{e^{{{i2\pi } \mathord{\left/ {\vphantom {{i2\pi } 5}} \right. \kern-\nulldelimiterspace} 5}}},{e^{{{i4\pi } \mathord{\left/ {\vphantom {{i4\pi } 5}} \right. \kern-\nulldelimiterspace} 5}}},{e^{{{i6\pi } \mathord{\left/ {\vphantom {{i6\pi } 5}} \right. \kern-\nulldelimiterspace} 5}}},{e^{{{i8\pi } \mathord{\left/ {\vphantom {{i8\pi } 5}} \right. \kern-\nulldelimiterspace} 5}}}$ i.e. $\alpha ,\beta ,\gamma ,\delta$

Step 2 of 2:

Now it is given that $\alpha ,\beta ,\gamma ,\delta$ \text{are roots of equation (1)

And for ${x^5} = 1$ , 5th root is 1

So, from the property of the nth root of unity. Suppose the positive value of the integer is the nth root of unity.

All the n roots of the nth roots of unity are in GP. If we add all the n roots of the nth root, the resulting value equals zero.

Therefore, ${1^{2021}} + {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {\delta ^{2021}} = 0$

$\Rightarrow {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {\delta ^{2021}} = - 1$

Final Answer:

Hence, Option (B) is correct.

Solution:

Step 1 of 2:

Step 2 of 2:

Final Answer:

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