If $x^2-x+1=0$, the value of  $\sum _{n=1}^5 {\left(x^n+\frac{1}{x^n}\right)}^2$ is

Observe that the roots of $x^2-x+1=0$ are $-\omega$ and $-{\omega }^2$

$\sum _{n=1}^5 {\left(x^n+\frac{1}{x^n}\right)}^2=\sum _{n=1}^5 {\left(x^{2n}+\frac{1}{x^{2n}}+2\right)}^{ }$

Taking $x=-\omega$

Required sum $=\left({\omega }^2+{\omega }^4+{\omega }^6+{\omega }^8+{\omega }^{10}\right)$

$\begin{array}{r}+\frac{1}{{\omega }^2}+\frac{1}{{\omega }^4}+\frac{1}{{\omega }^6}+\frac{1}{{\omega }^8}+\frac{1}{{\omega }^{10}}\\ =\left({\omega }^2+\omega +1+{\omega }^2+\omega \right)\\ +\left(\omega +{\omega }^2+1+\omega +{\omega }^2\right)+10\\ =(-1)+(-1)+10=8\end{array}$