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$