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

We need to find $$ \alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021} $$, where $$ \alpha, \beta, \gamma, \delta $$ are roots of $$ x^4 + x^3 + x^2 + x + 1 = 0 $$.

The equation $$ x^4 + x^3 + x^2 + x + 1 = 0 $$ can be written as:

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

So the roots are the primitive 5th roots of unity: $$ \omega, \omega^2, \omega^3, \omega^4 $$, where $$ \omega = e^{2\pi i/5} $$.

Each root satisfies $$ \omega^5 = 1 $$.

$$2021 = 5 \times 404 + 1$$

Therefore, $$ 2021 \mod 5 = 1 $$.

For each root $$ \omega^k $$ (where $$ k = 1, 2, 3, 4 $$):

$$(\omega^k)^{2021} = (\omega^k)^{5 \times 404 + 1} = ((\omega^k)^5)^{404} \cdot \omega^k = 1^{404} \cdot \omega^k = \omega^k$$

$$\alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021} = \omega + \omega^2 + \omega^3 + \omega^4$$

By Vieta's formulas for $$ x^4 + x^3 + x^2 + x + 1 = 0 $$, the sum of roots equals $$ -1 $$ (negative of the coefficient of $$ x^3 $$).

$$\omega + \omega^2 + \omega^3 + \omega^4 = -1$$

The answer is $$ -1 $$, which corresponds to Option D.