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

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