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 $\Bbb C$?

A bit late answer but I think worth mentioning it.
Here a direct calculation which uses only once the fact


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*$\frac{2}{x}=2-x \stackrel{x\neq 0}{\Leftrightarrow} \boxed{x^2 = 2x-2} \quad (\star)$
\begin{eqnarray*} \left[x^9-(x^4+x^2+1)(x^6+x^3+1)\right]^3
& = & \left[x^9-\frac{x^6-1}{x^2-1}\frac{x^9-1}{x^3-1}\right]^3\\
& = & \left[x^9-\frac{x^3+1}{x^2-1}(x^9-1)\right]^3\\
& = & \left[x^9-\frac{x^2-x+1}{x-1}(x^9-1)\right]^3 \\
& \stackrel{(\star)}{=} & \left[x^9-\frac{2x-2-x+1}{x-1}(x^9-1)\right]^3 \\
& = & \left[x^9-(x^9-1)\right]^3 \\
& = & 1 \\
\end{eqnarray*}
