If $$x^2 + x + 1 = 0$$, then $$x^{2018} + x^{2019}$$ equals which of the following:

We know that,

$$x^{3} - 1 = (x - 1)(x^{2} + x + 1)$$

Since, $$x^2 + x + 1 = 0$$

$$\therefore $$ $$x^{3} - 1$$ = 0 

=> $$x^{3}$$ = 1 

Now, x=1 is a solution but its not the only solution, as there are 2 more solutions which are not real numbers. Hence we cant simply substitute x=1 and obtain the value as 2. however we can substitute $$x^3=1$$ because cubinig the other 2 complex solutions will also give us 1. so we use $$x^3=1$$ and simply the expression . 

Now, $$x^{2018} + x^{2019}$$ 

= $$(x^{3})^{672} * x^{2}$$ + $$(x^{3})^{673}$$

= $$1^{672} * x^{2}$$ + $$1^{673}$$

= $$x^{2}$$ + 1

= -x

Hence, option C.