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.

Hint: $$x^2+\frac{1}{x^2} = 1 \quad \Longrightarrow \quad x^4 = x^2-1 \quad \Longrightarrow \quad x^6=x^2x^4=-1.$$
