Question

If $x+\frac{1}{x}=1$ ,then find the value of $x^{2000}+\frac{1}{x^{2000}}$.

Answer 1

Given $x+\frac{1}{x}=1$

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

Roots are $-\omega ,-{\omega }^2$

Here $\omega$ is cube root of unity.

$\Rightarrow 1+\omega +{\omega }^2=0$

${\omega }^3=1$

So,

$x^{2000}+\frac{1}{x^{2000}}\Rightarrow$ taking $x=4$

$\Rightarrow {\omega }^{2000}+({\omega }^2{)}^{2000}$

$({\omega }^3{)}^{666}{\omega }^2+({\omega }^3{)}^{1333}\left(\omega \right)\Rightarrow {\omega }^2+\omega$

$\therefore {\omega }^2+\omega =-1$.