calculate $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^{6}+1$ given $(x+x^{-1})^2 = 3$ If $\left(x+\dfrac 1 x\right)^2=3$ then the value of $$x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^{6}+1.$$
I'm trying to solve it like this $$x^2+\dfrac {1}{x^2}=1\text{ and }; x^6+\dfrac {1}{x^6}=-2$$
then solve the expression like this
$$(x^6+1)(x^{200}+x^{84}+x^{12}+1).$$
from here I am stuck.
Thanks in advance.

Since, we have
$x^2=x^4+1 \implies x^4=x^2-1 \implies x^8 = -x^2$.
Then, we get
$x^6 = x^2 .x^4= -1 $
$x^{12} = x^6 . x^6 = 1$
$x^{18} = x^{12}.x^6= (1)(-1)=-1$
$x^{84}= x^{(18).(4)}.x^{12}=(-1)^4 .(1)=1$
$x^{90}=x^{(18)(5)}=(-1)^5=-1$
$x^{200}=x^{(12)(16)}.x^6.x^2=(1)^{16}.(-1).x^2=-x^2$
$x^{206}=x^{200}.x^6=(-x^2).(-1)=x^2$
Therefore,
$x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1=x^2-x^2 -1+1-1+1-1+1=0.$
