if x²+ x+ 1=0 then find the value of

summation_r=1 ²⁷(x^r+ 1/x^r) ² .

let us consider the equation $x^3-1=0$ ...........(1)
the roots of the eq(1) are the cube roots of unity
$x^3-1=(x-1)(x^2+x+1)$
the imaginary roots $\omega and {\omega }^2$ are the roots of the eq $x^2+x+1=0$
now let $x=\omega$ ,
$and let P={\omega }^r+\frac{1}{{\omega }^r}$
here r can be either of 3k, 3k+1 or 3k+2 type
if r = 3k
$$\begin{gathered} P={\omega }^{3k}+\frac{1}{{\omega }^{3k}}=({\omega }^3{)}^k+\frac{1}{({\omega }^3{)}^k} \\ =1+1 \\ =2 \end{gathered}$$
if r = 3k+1
$$\begin{gathered} P={\omega }^{3k+1}+\frac{1}{{\omega }^{3k+1}} \\ =\omega .({\omega }^3{)}^k+\frac{1}{\omega .({\omega }^3{)}^k} \\ =\omega +\frac{1}{\omega } \\ =\omega +{\omega }^2 [\frac{1}{\omega }=\frac{{\omega }^2}{{\omega }^3}={\omega }^2 \\ =-1 \end{gathered}$$
if r = 3k+2
$$\begin{gathered} P={\omega }^{3k+2}+\frac{1}{{\omega }^{3k+2}} \\ ={\omega }^2.({\omega }^3{)}^k+\frac{1}{{\omega }^2.({\omega }^3{)}^k} \\ ={\omega }^2+\frac{1}{{\omega }^2}={\omega }^2+\omega [\sin ce \frac{1}{{\omega }^2}=\frac{\omega }{{\omega }^3}=\omega \\ P=-1 \end{gathered}$$
therefore
$$\begin{gathered} {\sum }_{r=1}^{27}{\left(x^r+\frac{1}{x^r}\right)}^2 \\ ={\sum }_{k=1}^9{\left(x^{3k}+\frac{1}{x^{3k}}\right)}^2+{\sum }_{k=0}^8{\left(x^{3k+1}+\frac{1}{x^{3k+1}}\right)}^2+{\sum }_{k=0}^8{\left(x^{3k+2}+\frac{1}{x^{3k+2}}\right)}^2 \\ ={\sum }_{k=1}^9(2{)}^2+{\sum }_{k=0}^8(-1{)}^2+{\sum }_{k=0}^8(-1{)}^2 \\ ={\sum }_{k=1}^{9}4+{\sum }_{k=0}^{8}1+{\sum }_{k=0}^{8}1 \\ =4*9+1*9+1*9 \\ =36+9+9 \\ =54 \end{gathered}$$

hope this helps you