Question

x²+x+1= 0 then x²⁰¹⁸+x²⁰¹⁹ is?

Answer 1

Step-by-step explanation:

here omega and omega square are roots

here w indicates omega

$1 + w + {w}^{2} = 0$

satisfies this equation

this is also known as cubeth roots of unity

conditions for these are

$1 + w + {w}^{2} = 0 \\ w^{3} = 1$

here the given equation satisfies for cubeth root of unity

$x = w \\ x = {w}^{2}$

now substituting them gives

${x}^{2018} + {x}^{2019} \\ {(w)}^{2018} + (w)^{2019} \\ = > 1 + w^{2}$

$= > - w = \frac{1 + \sqrt{3}i }{2}$

here is the answer