Let $$\alpha$$ be a root of the equation $$1 + x^2 + x^4 = 0$$. Then the value of $$\alpha^{1011} + \alpha^{2022} - \alpha^{3033}$$ is equal to:

$$\alpha$$ is a root of $$1 + x^2 + x^4 = 0$$. We wish to find $$\alpha^{1011} + \alpha^{2022} - \alpha^{3033}$$.

Let $$y = x^2$$ so that the equation becomes $$y^2 + y + 1 = 0$$. Its roots are $$y = \omega$$ and $$y = \omega^2$$, where $$\omega = e^{2\pi i/3}$$ is a primitive cube root of unity satisfying $$\omega^3 = 1$$.

Since $$\alpha^2 = \omega$$ (or $$\alpha^2 = \omega^2$$), it follows that $$\alpha^6 = (\alpha^2)^3 = \omega^3 = 1$$, so $$\alpha$$ is a primitive 6th root of unity.

Reducing exponents modulo 6, we write $$1011 = 168\times6 + 3$$ so $$\alpha^{1011} = \alpha^3$$, $$2022 = 337\times6 + 0$$ so $$\alpha^{2022} = \alpha^0 = 1$$, and $$3033 = 505\times6 + 3$$ so $$\alpha^{3033} = \alpha^3$$.

Therefore, $$\alpha^{1011} + \alpha^{2022} - \alpha^{3033} = \alpha^3 + 1 - \alpha^3 = 1$$. Option A: $$1$$