<s>[INST] 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.
 [/INST] 
$$\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$

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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