If $x^2+x+1=0,$  then the value of ${(x+1/x)}^2+{(x^2+1/x^2)}^2$  $+\mathrm{...}+{(x^{27}+1/x^{27})}^2$  is

$x^2+x+1=0\Rightarrow x=\omega$  or ${\omega }^2$  $\left( \sin ce 1+\omega +{\omega }^2=0\right)$

Let $x=\omega .$  Then,

$x+\frac{1}{x}=\omega +\frac{1}{\omega }=\omega +{\omega }^2=-1$

$x^2+\frac{1}{x^2}={\omega }^2+\frac{1}{{\omega }^2}={\omega }^2+\omega =-1$          $\left(\sin ce \frac{1}{\omega }={\omega }^2\right)$

$x^3+\frac{1}{x^3}={\omega }^3+\frac{1}{{\omega }^3}=2$

$x^4+\frac{1}{x^4}={\omega }^4+\frac{1}{{\omega }^4}=\omega +{\omega }^2=-1.etc.$

$\therefore {\left(x+\frac{1}{x}\right)}^2+{\left(x^2+\frac{1}{x^2}\right)}^2+{\left(x^3+\frac{1}{x^3}\right)}^2+\mathrm{...}+{\left(x^{27}+\frac{1}{x^{27}}\right)}^2$

$=\left[{\left(x+\frac{1}{x}\right)}^2+{\left(x^2+\frac{1}{x^2}\right)}^2+{\left(x^4+\frac{1}{x^4}\right)}^2+\mathrm{...}+{\left(x^{26}+\frac{1}{x^{26}}\right)}^2\right]$

$+[{\left(x^3+\frac{1}{x^3}\right)}^2+{\left(x^6+\frac{1}{x^6}\right)}^2+\left(x^9+\frac{1}{x^9}\right)$ $\mathrm{...}+{\left(x^{27}+\frac{1}{x^{27}}\right)}^2$

=$\left[{\left(-1\right)}^2+{\left(-1\right)}^2+.....+{\left(-1\right)}^2\right]$ $+\left(2^2+2^2+......+2^2\right)$

$=18+9\left(2^2\right)=54$