Question

If $\frac{x+1}{x}=1$, then $x^{200}+\frac{1}{x^{200}}$ equals

• A. $-1$
• B. $0$
• C. $1$
• D. ${\omega }^2$

Answer 1

Answer: (A)

$x+1/x=1$
$\Rightarrow x^2-x+1=0$
$\Rightarrow (x+\omega )(x+{\omega }^2)=0($ where $\omega =\frac{-1+i\sqrt{3}}{2})$
$\Rightarrow x=-\omega ,-{\omega }^2$
when $x=-\omega$ then $x^{200}+\frac{1}{x^{200}}={\omega }^{200}+\frac{1}{{\omega }^{200}}$
$=({\omega }^3{)}^{66}\cdot {\omega }^2+\frac{1}{({\omega }^3{)}^{66}\cdot {\omega }^2}=-1$
$(\because {\omega }^3=1)$
$x=-{\omega }^2$ then $x^{200}+\frac{1}{x^{200}}=(-{\omega }^2{)}^{200}+{\left(\frac{1}{-{\omega }^2}\right)}^{200}$
$={\omega }^{400}+\frac{1}{{\omega }^{400}}=\omega +{\omega }^2=-1$