<s>[INST] 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 $\Bbb C$?
 [/INST] 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.
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