Try everything

$\begin{aligned} x^2 + x + 1 & = 0 \quad \quad \small \color{#3D99F6}{\text{Multiply both sides by }x} \\ x^3 + x^2 + x & = 0 \quad \quad \small \color{#3D99F6}{\text{Add 1 both sides}} \\ x^3 + \color{#3D99F6}{x^2 + x + 1} & = 0 + 1 \\ x^3 + \color{#3D99F6}{0} & = 1 \\ \implies x^3 & = 1 \end{aligned}$

Also we have:

$\begin{aligned} x^2 + x + 1 & = 0 \\ x + 1 + \frac{1}{x} & = 0 \\ x + \frac{1}{x} & = -1 \\ x^2 + \frac{1}{x^2} & = \left(x + \frac{1}{x}\right)^2 - 2 = - 1 \\ x^3 + \frac{1}{x^3} & = 1 + 1 = 2 \end{aligned}$

Since $x^3 = 1\implies x^n + \dfrac{1}{x^n} = \begin{cases} 2 & \text{if } n \equiv 0 \text{ (mod 3)} \\ -1 & \text{if } n \not{\equiv} 0 \text{ (mod 3)} \end{cases}$

$\begin{aligned} S & = \left(x + \frac{1}{x}\right)^2 + \left(x^2 + \frac{1}{x^2}\right)^2 + \left(x^3 + \frac{1}{x^3}\right)^2 + \cdots + \left(x^{2016} + \frac{1}{x^{2016}}\right)^2 \\ & = 1+1+4+1+1+4+\cdots+4 \\ & = 2016(1) + (4-1)\left \lfloor \frac{2016}{3} \right \rfloor \\ & = 2016 + 2016 \\ & = \boxed{4032} \end{aligned}$

Clearly , We know by cube roots of unity,.....

That x= omega , (omega)^2,.......

Now, evaluating the sum of first 3 expansions (lets call it a triad!!), by using the rules of cube roots of unity,

We get →

1 + 1 + 4 = 6

Now this triad....keeps on repeating itself and hence appears 2016/3 =672 times ......therefore required sum→

672 × 6 = 4032...{ans.}

$x^2+x+1=0$ implies that $x+\frac{1}{x}=-1$ .

If $\theta = \frac{2\pi}{3}$ and $x=\cos\theta+i\sin\theta$ , then $x+\frac{1}{x}=2\cos\theta=-1$ .

Now, $x^n+\frac{1}{x^n}=2\cos n\theta=2\cos\frac{2n\pi}{3} = \begin{cases} \pm 2 & \text{if } n=3k \\ \pm 1 & \text{if } n \neq 3k\end{cases}$ . So $\left(x^n+\frac{1}{x^n}\right)^2= \begin{cases} 4 & \text{if } n=3k \\ 1 & \text{if } n \neq 3k\end{cases}$ .

Hence the desired answer is $(1+1+4) \times \frac{2016}{3}=4032$ .

Take x = w(omega) So x+1/x = -1 But x^3n = 1 Therefore Answer( 1+1+4)*2016/3

$Let f(n)=X^n+\dfrac 1 {X^n}.\\ X^2+X+1=0,\ \ \implies\ \ x+\dfrac 1 X= - 1,\ \ \implies\ f(1)= -1. ...\color{#3D99F6}{**}\ \ \ \ \ \ \therefore\ \ \color{#3D99F6}{\{f(1)\}^2=1}\\ f(2)+2=f(1)^2,\ \ \implies\ \ f(2)=-1.\ \ \ \therefore\ \ \color{#3D99F6}{\{f(2)\}^2=1}\\ f(3)+3*1* f(1)=f(1)^3= - 1,\ \ \implies\ f(3)=2\ \ \ \ \ \ \therefore\ \ \color{#3D99F6}{\{f(3)\}^2=4}\\ \{f(2)\}^2=f(4)+2=1,\ \ \ \ \implies\ f(4)=-1\ \ \ \ \ \ \therefore\ \ \color{#3D99F6}{\{f(4)\}^2=1}\\ f(2)*f(3)=- 2,\ \ but\ \ f(2)*f(3)=f(5) - f(1)=f(5) - (- 1) \ \ \ \ \ \ \therefore\ \ \color{#3D99F6}{\{f(5)\}^2=1}\\ f(3)^2=f(6)+2,\ \ but\ \ f(3)^2=4, \ \ \ \ \ \ \therefore\ \ \color{#3D99F6}{\{f(6)\}^2=4}\\ \therefore\ \ this \ is\ cyclic\ function\ with \ period\ 3.\\ \therefore\ \ f(\ n \equiv 1\ or\ 2\ ( mod\ 3)\ )^2=1, \ and\ \ \ \ f(\ n\ \equiv\ 0\ (mod\ 3)\ )=4\\ \implies\ each \ cycle\ of\ 3\ adds\ 6.\\ 3|2016,\ \ \therefore\ \ Total=2*2016=\color{#D61F06}{4032}$

Just take x as a cube root of unity . Then its easy