Question

If $x^{2}-x+1=0$, then find value of $\displaystyle\sum_{n=1}^{5}\left(x^{n}+\frac{1}{x^{n}}\right)^{2}$.

Answer 1

$x^{2}-x+1=0$
$\Rightarrow x=\frac{1 \pm \sqrt{3} i}{2}=-\omega,-\omega^{2}$
$\therefore \displaystyle\sum_{n=1}^{5}\left(x^{2 n}+\frac{1}{x^{2 n}}+2\right)$
$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}+2\right)+\left(x^{4}+\frac{1}{x^{4}}+2\right)+$
$\left(x^{6}+\frac{1}{x^{6}}+2\right)+\left(x^{8}+\frac{1}{x^{8}}+2\right)+$
$\left(x^{10}+\frac{1}{x^{10}}+2\right)$
$\Rightarrow\left(\omega^{2}+\omega^{4}+\omega^{6}+\omega^{8}+\omega^{10}\right)+$
$\left(\frac{1}{\omega^{2}}+\frac{1}{\omega^{4}}+\frac{1}{\omega^{6}}+\frac{1}{\omega^{8}}+\frac{1}{\omega^{10}}\right)+10$
$\Rightarrow-1-1+10=8 .$