If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$

Hint

$$x^2=2x-2$$

$$x^3=2x^2-2x=2(2x-2)-2x=2x-4$$

$$x^4=(2x-2)^2=4-8x+4(2x-2)=-4$$

$$x^6=-4(2x-2)=?$$

$$x^9=(-4)^2x=?$$

It's enough to prove that $$x^9-(x^4+x^2+1)(x^6+x^3+1)=1$$ or $$(x^3-1-(x^4+x^2+1))(x^6+x^3+1)=0,$$ for which it's enough to prove that $$x^4-x^3+x^2+2=0$$ or $$x^4-2x^3+2x^2+x^3-2x^2+2x+x^2-2x+2=0$$ or $$(x^2-2x+2)(x^2+x+1)=0,$$ which is obvious.