# Solve the equation $x^6-2x^5+3x^4-3x^2+2x-1=0$

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.

• I'd note x=1 is a solution, then divide out by (x-1) using synthetic division
– Alan
Aug 31, 2022 at 22:45
• and -1 is a solution Aug 31, 2022 at 22:47
• 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 Aug 31, 2022 at 22:47

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 :)

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

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.

• very elegant, thanks Aug 31, 2022 at 22:56
• 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. Sep 1, 2022 at 3:17

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 $$\dfrac{1 \pm \sqrt{(1-i)^2-4}}{2}$$