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Solve the equation $$x^6-2x^5+3x^4-3x^2+2x-1=0$$ Let's divide both sides of the equation by $x^3\ne0$ (as $x=0$ is obviously not a solution, we can consider $x\ne0$). Then we have $$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$ What do we do now? If we say $y=x-\dfrac{1}{x}$, we won't be able to express $\left(x^2-\dfrac{1}{x^2}\right)$ in terms of $y$ because of the minus sign as $(a-b)^2=a^2-2ab\color{red}{+b^2}$. On the other side, $$y^3=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right)\\x^3-\dfrac{1}{x^3}=y^3+3y$$ I think this is a traditional issue when solving reciprocal equations, but I can't figure out how to deal with it.

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    $\begingroup$ I'd note x=1 is a solution, then divide out by (x-1) using synthetic division $\endgroup$
    – Alan
    Commented Aug 31, 2022 at 22:45
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    $\begingroup$ and -1 is a solution $\endgroup$
    – Mabadai
    Commented Aug 31, 2022 at 22:47
  • $\begingroup$ but watch its expansion, it looks like taylor, polynom and pascal triangle. If you compare the sixth power with the constant, and the odd powers (with -2 and 2), and the even powers(with +3 and -3). Then there is a pattern $\endgroup$
    – Mabadai
    Commented Aug 31, 2022 at 22:47

3 Answers 3

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Noticing that $x=1$ and $x=-1$ are roots and by long division, you'll get $x^6-2x^5+3x^4-3x^2+2x-1=(x-1)(x+1)(x^4-2x^3+4x^2-2x+1)$

Considering $x^4-2x^3+4x^2-2x+1=0$, you now divide by $x^2$ and you'll get $x^2+\frac{1}{x^2}-2(x+\frac{1}{x})+4=0$ which, using the substitution $y=x+\frac{1}{x}$, gives $y^2-2y+2=0$.

I believe you can finish this from here :)

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$$\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$

There are two roots $\pm1$ associated with the factor of $x-1/x$. To find the other roots, divide by $x-1/x$ to get

$$ x^2+1+{1\over x^2}-2\left(x+{1\over x}\right)+3=0 $$

Noticing that $(x+1/x)^2=x^2+2+1/x^2$ we have $$ (x+1/x)^2-2(x+1/x)+2=0, $$ This quadratic has solutions $x+1/x=1\pm i$. Then you need to solve $x^2-(1\pm i)x+1=0$ to get the final roots.

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  • $\begingroup$ very elegant, thanks $\endgroup$
    – Mabadai
    Commented Aug 31, 2022 at 22:56
  • $\begingroup$ I tried to do this problem without paper, just from typing latex, and there were so many errors, and each time I looked at it again I found another error. I had to do some job, then I came back, and it was really annoying me that there were so many errors, so I did it all on a piece of paper, got the solution in a couple of minutes. I suppose the lesson here is that I can't type latex and do mathematics problems at the same time, or something. $\endgroup$ Commented Sep 1, 2022 at 3:17
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The equation is:-
$$x^6-2x^5+3x^4-3x^2+2x-1=0$$
Continuing the given method we get:-
$$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$ $$\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+1\right)-2\left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)+3\left(x-\dfrac{1}{x}\right)=0$$ $$\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+4\right)=0$$ Thus we see $\left(x-\dfrac{1}{x}\right)=0$ is a solution.
So $$x^2-1=0\\\Rightarrow x^2=1\\\Rightarrow x=\pm1$$ On the other hand:- $$\left(x^2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+4\right)=0$$ Denoting $x+\dfrac{1}{x}$ as $a$ the equation becomes:- $$a^2-2a+2=0$$ Thus $$a = \dfrac{2 \pm \sqrt{4-4\cdot2}}{2}$$ $$\Rightarrow x+\dfrac{1}{x} = \dfrac{2 \pm 2i}{2}$$ $$\Rightarrow \dfrac{x^2+1}{x} = {1 \pm i}$$ Thus we get 2 equations:- $x^2-(1+i)x+1=0$ and $x^2-(1-i)x+1=0$
Thus $$x = \dfrac{1 \pm \sqrt{(1+i)^2-4}}{2}\\x = \dfrac{1 \pm \sqrt{(1-i)^2-4}}{2}$$ Thus the solutions to the equation are:-
$\pm1$,$\dfrac{1 \pm \sqrt{(1+i)^2-4}}{2}$ and $\dfrac{1 \pm \sqrt{(1-i)^2-4}}{2}$

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