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

Question

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 $\Bbb C$?

Answer 1

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.

Answer 2

Hint

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

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

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

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

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

Answer 3

A bit late answer but I think worth mentioning it.

Here a direct calculation which uses only once the fact

• $\frac{2}{x}=2-x \stackrel{x\neq 0}{\Leftrightarrow} \boxed{x^2 = 2x-2} \quad (\star)$

\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*}