Given equation is $x^{2}+x+1=0$
$\Rightarrow \quad x=\omega$ and $x=\omega^{2}$
Case I : When $x=\omega$
Then
$\sum_{n=1}^{6}\left[x^{n}+\frac{1}{x^{n}}\right]^{2}=\sum_{n=1}^{6}\left[\omega^{n}+\omega^{2 n}\right]^{2}\left[\because \frac{1}{\omega}=\omega^{2}\right]$
$=\left(\omega+\omega^{2}\right)^{2}+\left(\omega^{2}+\omega^{4}\right)^{2}+\left(\omega^{3}+\omega^{6}\right)^{2}$
$+\left(\omega^{4}+\omega^{8}\right)^{2}+\left(\omega^{5}+\omega^{10}\right)^{2}+\left(\omega^{6}+\omega^{12}\right)^{2}$
$\begin{array}{r}
=(-1)^{2}+(-1)^{2}+(2)^{2}+(-1)^{2} \\
+(-1)^{2}+(2)^{2}=12
\end{array}$
Case II: When $x=\omega^{2}$
Then
$\begin{array}{c}
\sum_{\mathrm{n}=1}^{6}\left[\mathrm{x}^{\mathrm{n}}+\frac{1}{\mathrm{x}^{\mathrm{n}}}\right]^{2}=\sum_{\mathrm{n}=1}^{6}\left[\omega^{2 \mathrm{n}}+\omega^{\mathrm{n}}\right]^{2}\left[\because \frac{1}{\omega^{2}}=\omega\right] \\
=12
\end{array}$