A precalculus solution for $x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 =0$ 
Using algebra (precalculus) and suggest the solution method for the polynomial $$x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 =0$$

I'm solving problems on polynomials.  I'm stuck here.
My attempts.
First, I tried the Rational root theorem, then I failed.
Then I tried factorise the polynomial e.g. $(x^2+ax+b)(x^4+cx^3+dx^2+ex+f)$, but I failed again.
At the end I tried
$$P(x)/x^3=x^3-3x+2+\frac 3x-3\frac {1}{x^2}+\frac {1}{x^3}=x^3+\frac {1}{x^3}-3\bigg(x-\frac 1x\bigg)-\frac {3}{x^2}+2=0$$
I failed again.
 A: Let,
$$P(x)=x^6-3x^4+2x^3+3x^2-3x+1=0$$
We see that $x=0$ is not one of the roots of $P(x)$. Therefore, we can divide all terms of the polynomial $P(x)$ by $x^2:$
$$
\begin{align}
\frac {P(x)}{x^2}&=\color{red}{x^4}-\color{blue}{3x^2}+\color{red}{2x}+\color{green}{3}-\color{blue}{\frac {3}{x}}+\color{red}{\frac {1}{x^2}}\\
&=\color{red}{x^4+2x+\frac {1}{x^2}}-\color{blue}{3\left(x^2+\frac 1x\right)}+\color{green}{3}\\
&=\left(x^2+\frac 1x\right)^2-3\left(x^2+\frac 1x\right)+3=0.\end{align}
$$
Then substitute $x^2+\frac 1x=u$ , we obtain:
$$
\begin{align}u^2-3u+3=0\\
\implies u_{1,2}=\frac{3\pm i\sqrt 3}{2}\end{align}
$$
If you want to find the roots by radicals in exact form, you will need to solve the following cubic equation :
$$x^2+\frac 1x=u\implies x^3-ux+1=0$$
where, $u\in\left\{u_1,u_2\right\}$.
Thus, you have shown that the polynomial $P(x)$ is solvable by radicals.
A: We can rewrite the given polynomial as
$$p(x) = x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 = \\
(x^6 + 2x^3 + 1) - 3x (x^3 - x + 1) = \\
(x^3 + 1)^2 - 2x (x^3 + 1) + 2x^2 - x (x^3 - x + 1) = \\
(x^3 - x + 1)^2 - x (x^3 - x + 1) + x^2.$$
Therefore, if $x$ is a root of $p$, then either
$$x^3 - x + 1 = -\omega x$$
or
$$x^3 - x + 1 = -\omega^2 x$$
where $\omega = e^{2\pi i / 3} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$ and $\omega^2 = \bar\omega = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$.
Since each of these is a cubic whose coefficients are expressible as radicals, the solutions to these individual cubics are also expressible as radicals.
(Honestly, though, the above sequence of steps was somewhat reverse-engineered from the eventual solution that I found based on numerical experimentation.  The route we came to this in the comments was: through using a CAS to calculate the Galois group of the polynomial, use that to conjecture what solutions in terms of radicals might look like, and then do numerical experimentation based on that form to try to find elements of the form.)
(Note also that we can see a relation of this solution to the CAS-computed Galois group of the polynomial.  The solution above gives the splitting field of $p$ as a compositum of two extensions of $\mathbb{Q}(\sqrt{-3})$, each of degree at most 6.  Assuming the CAS is correct that the Galois group has order 72, that implies that the two extensions have minimal intersection $\mathbb{Q}(\sqrt{-3})$ and each have degree 6.)
A: Notice that the only polynomials that are solvable by radicals for arbitrary coefficients are those of degree less than or equal to $3$, i.e cubics. Also, notice that the equation of question is of degree $6$, so why not try to write it as the product of two cubics? If this is possible, then solutions can be written using radicals by solving both cubics.
$$(x^3+\alpha_1x^2+\alpha_2x+\alpha_3)(x^3+\alpha_4x^2+\alpha_5x+\alpha_6)$$
If you expand the product, you can deduce that $\alpha_4=-\alpha_1$ because the coefficient of $x^5$ is $0$. You can also deduce $\alpha_3\alpha_6=1$. Furthermore, you will get $4$ equations in $5$ unknowns, meaning there is a free variable, one unknown that we can choose freely if we are just trying to find a particular solution, which we are in that case. The best value to set for a variable would be $0$ since it would eliminate terms and thus, we get a simpler system. Since we have $\alpha_3\alpha_6=1$, neither can be $0$. Since $\alpha_1$, as you would be able to see if you expand, appears in $2$ terms in two equations, we can set it to $0$ since it simplifies the system of equations the most. The rest is solving quadratics and taking reciprocals. As a result, you will get the solution (product of two reduced cubics):
$$\left(x^3+\frac{-3\mp_1 i\sqrt 3}{9}x\pm_2\sqrt 2+1\right)\left(x^3+\frac{-3\pm_1 i\sqrt 3}{2}x\pm_2\sqrt 2-1\right)$$
where the $\pm$'s with the same subscript are dependent. For example, for $\pm_1$, if the first term takes the $+$ branch, the other takes the $-$ branch.
