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.

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.
