QJEEM22Maaae7: JEE Main 2022 Mathematics - Class 11 JEEM22Maaae7 Solution: If are the roots of the equation , then is equal to :

When, ${x^5} = 1$

then ${x^5} - 1 = 0$

$\Rightarrow (x - 1)({x^4} + {x^3} + {x^2} + x + 1) = 0$

Given, ${x^4} + {x^3} + {x^2} + x + 1 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ and 8.

$\therefore$ Roots of ${x^5} - 1 = 0$ are 1, $\alpha$ , $\beta$ , $\gamma$ and 8.

We know, Sum of p^th power of n^th roots of unity = 0. (If p is not multiple of n) or n (If p is multiple of n)

$\therefore$ Here, Sum of p^th power of n^th roots of unity

$$=1^p+{\alpha }^p+{\beta }^p+{\gamma }^p+8^p=\{\begin{array}{ccc}0& ;& \mathrm{If}\mathrm{p}\mathrm{is}\mathrm{not}\mathrm{multiple}\mathrm{of}5\\ 5& ;& \mathrm{If}\mathrm{p}\mathrm{is}\mathrm{multiple}\mathrm{of}5\end{array}$$

Here, $p = 2021$ , which is not multiple of 5.

$\therefore$ ${1^{2021}} + {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {8^{2021}} = 0$

$\Rightarrow {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {8^{2021}} = - 1$