Brain works 4

My solution is similar to Tanishq's but more elementary as I'm not explicitly using complex numbers. Observe that $(x^2+x+1)(x-1)=x^3-1=0$ , so that $x^3=1$ . It follows that $x^{3k}=1,x^{3k+1}=x,x^{3k+2}=x^2$ for all (positive or negative) integers $k$ . Thus $x^m+\frac{1}{x^m}=1+1=2$ when $m$ is divisible by 3 and $x^m+\frac{1}{x^m}=x+x^2=-1$ otherwise. Since there are 9 summands of the first kind (with $m$ divible by 3) and 18 of the latter, the answer is $9\times2^2+18(-1)^2=54$ .

$x^2+x+1=(x-\omega)(x-\omega^2)$ so $x$ is $\omega~or ~\omega^2$

$\omega=\frac{1}{\omega^2}$ and $\frac{1}{\omega}=\omega^2$

also $1+\omega +\omega^2 =0$ and $\omega+\omega^2=-1$

also $\omega^{3r}=1$ so $\omega^{3r+1}=\omega$

$(\omega+ \frac{1}{\omega})^2+(\omega^2 +\frac{1}{\omega^4})^2 +(\omega^3+\frac{1}{\omega^6})^2+ ...................$

we observe terms of form $3r+1$ and $3r+2$ and $3r$ are 9 each.(i am talking about power of $x$ i.e $x,x^2,x^3,x^4,x^5....$ )

terms of form $3r+1$ have value $1$ . similarly terms of form $3r+2$ have value $1$ and terms of form $3r$ have value equal to $2^2$ .

so answer is $9+9+(2)^{2}\times 9=54$

We will use the fact that if x+1/x=2cosA then x^n + 1/x^n = 2 cos(nA)

Since x^2 + x + 1=0 this implies x+1/x=-1=2cos(2pi/3)

Required answer = 2^2((cos(2pi/3))^2 + (cos(4pi/3))^2+ (cos(6pi/3))^2)*9 =54

$x^2+x+1=0~\implies~x+\dfrac 1 x=-1.~ Let~~ f(n)=x^n+\dfrac{ 1 }{x^n}\\We ~observe ~that~ f\{n \equiv 1 \pmod3 \}= - 1. ~~ f\{n \equiv 2 \pmod3 \}= - 1. \\But~ f\{n \equiv 0 \pmod3 \}= 2. ~~~~~~~~~\therefore~f(n)^2~are,~1,~1,~4,~respectively.\\For~~ n=1,2,3, . . .27,~~there~ are~\dfrac{27}{3}=9~of~ each~of~the~three~types.\\\therefore~the~SUMATION=9*1+9*1+9*4=\color{#D61F06}{\boxed{54}}$