Cube Me!

$x+\frac1x=-1\Rightarrow x^2+x+1=0$

$(x-1)(x^2+x+1)=0$

$x^3-1=0\Rightarrow x^{3k}\equiv 1$

$x^{1999}+\frac{1}{x^{1999}}=x\cdot (x^3)^{666}+\frac{1}{x\cdot(x^3)^{666}}=x+\frac1x=-1$

For the equation x+1/x=-1, Solutions are imaginary cubic roots of 1 = ω and ω²

also ω³=1, if substitue ω or ω² in the below equation, you will get -1.

x^1999 + 1/x^1999= -1 if x= ω , ω²

The solutions of the equation x+1/x=-1 are ω and ω² (the complex cubic roots of

unity) where ω³=1

x^1999 + 1/x^1999 = x . (x^3)^666 + 1/x . (x^3)^666 = x+1/x=-1