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

$$ = {1^p} + {\alpha ^p} + {\beta ^p} + {\gamma ^p} + {8^p} = \left\{ {\matrix{ 0 & ; & {\mathrm{If\,p\,is\,not\,multiple\,of\,5}} \cr 5 & ; & {\mathrm{If\,p\,is\,multiple\,of\,5}} \cr } } \right.$$

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

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

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