<s>[INST] 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.
 [/INST] 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.)
</s>