If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$ If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$
After we solve for $x$ (its a quadratic), and find that $x=1\pm i$, it's trivial to see the powers of $x$ in the complex plane, but the problem must be solved without using the complex numbers. With complex numbers i found that the result is $1$.
Is there any way to solve this in $\Bbb R$ without entering $\Bbb C$?

Hint
$$x^2=2x-2$$
$$x^3=2x^2-2x=2(2x-2)-2x=2x-4$$
$$x^4=(2x-2)^2=4-8x+4(2x-2)=-4$$
$$x^6=-4(2x-2)=?$$
$$x^9=(-4)^2x=?$$
