If α , β , γ , δ are the roots of the equation x 4 + x 3 + x 2 + x + 1 = 0 , then α…

We know that $x^4+x^3+x^2+x+1=0$ can be written as 

$\frac{x^5-1}{x-1}=0$$\Rightarrow x^5=1$

i.e. $x={\left(1\right)}^{\frac{1}{5}}$

So $\alpha , \beta , \gamma$ and $\delta$ are the $5^{\text{th }}$ roots of unity except $1$.

Now, from the concept of roots of unity, we know

 ${\alpha }^n+{\beta }^n+{\gamma }^n+{\delta }^n+1^n=\left\{\begin{array}{ll}0& \text{if} n \text{is not a multiple of} 5\\ 5& \text{if} n \text{is a multiple of} 5\end{array}\right.$

So, ${\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}=-1$.