$a$,$b$ and $c$ are roots of the equation $x^3-x^2-x-1=0$ The roots of the equation $x^3-x^2-x-1=0$ are $a$,$b$ and $c$.
if $n \gt 21 $ and $n \in \mathbb{N}$ The find the possible values of $$E=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$$ in $[0 \: 2]$ are?
Since $x^3=x^2+x+1$ from the graphs of $x^3$ and $x^2+x+1$ its clear that they meet at one point. So number of real roots of $x^3-x^2-x-1=0$ is one which is positive and remaining two are complex conjugates. So let the roots be $a$ and $b=re^{i\theta}$ and $c=re^{-i\theta}$ But product of the roots is $$ar^2=1$$ hence $$r=\frac{1}{\sqrt{a}}$$
Now assuming $b=z$ and $c=\bar{z}$ then 
$$E=\frac{a^n-z^n}{a-z}+\frac{z^n-\bar{z}^n}{z-\bar{z}}+\frac{\bar{z}^n-a^n}{\bar{z}-a}$$ so
$$E=2 \Re\left(\frac{a^n-z^n}{a-z}\right)+r^{n-1}\frac{\sin(n\theta)}{\sin\theta}$$
I am not able to proceed from here

HINT: If 
$E_n=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$ 
then, because of $x^3=x^2+x+1$ you have $x^n=x^{n-1}+x^{n-2}+x^{n-3}$ and:
$E_n=E_{n-1} + E_{n-2} + E_{n-3}$. 
It is easy to find out the first values of this recurrence.
