Your polynomial is a reciprocal polynomial, in that its coefficients read the same forwards to backwards. A consequence of this is that once you restrict to $x \neq 0$ (which is not a root) and divide by $x^3$, you get $$
x^3 + \frac 1{x^3} -\left(x^2+\frac 1{x^2}\right) + 4\left(x+\frac 1x\right) - 4 = 0
$$
Use the high school tricks given by $$
x^2+\frac 1{x^2} = \left(x+\frac 1x\right)^2 - 2 \\
x^3+\frac 1{x^3} = \left(x+\frac 1x\right)^3 - 3 \left(x+\frac 1x\right)
$$
To get : if $z = x+\frac 1x$, then $$
(z^3-3z) - (z^2-2) + 4z - 4 = 0 \implies z^3-z^2+z-2 = 0
$$
Can this be solved using simple methods? The answer is no. This is a cubic polynomial : if it has to be reducible over the rationals, then it must have a linear factor. The rational root theorem fails, as one sees by trying the roots $\pm 1 , \pm 2$. Therefore, this polynomial is irreducible.
The only real root of this equation (that there is only one root can be found through calculus, for example) can be found explicitly through Cardano's method. It is given by
$$
z = \frac 13\left(1 - 2 \sqrt[3]{\frac{2}{47+3 \sqrt{249}}} + \sqrt[3]{\frac 12(47 + 3\sqrt{249})}\right)
$$
There are two other strictly complex roots $z_1$ and $\bar{z_1}$ (the complex conjugate of $z_1$) which drop out from this equation. They can be expressed in terms of $z$ as follows : $z_1\bar{z_1} = \frac 2{z}$ and $z_1+\bar{z_1} = 1-z$. This will tell you that $|z_1| = \sqrt{\frac{2}{z}}$ and $\text{Re}(z_1) = \frac{1-z}{2}$, so you can retrieve the complex part of $z_1$ and obtain an expression for $z_1$ entirely in terms of $z$.
Once you do this, you have three equations $$
x+\frac 1x = z , x+\frac 1x = z_1 , x+\frac 1x = \bar{z_1}
$$
which can be solved as ordinary quadratic equations to yield $$
x = \frac{1}{2}\left(z\pm \sqrt{z^2-4}\right) \\
x = \frac{1}{2}\left(z_1\pm \sqrt{z_1^2-4}\right) \\
x = \frac{1}{2}\left(\bar{z_1}\pm \sqrt{\bar{z_1}^2-4}\right)
$$
which are the six distinct roots of this problem. Note that all this detail is perhaps unimportant in computation : but it is useful if one is trying to compute relations between the roots, which is useful for computing Galois groups of these polynomials.
EDIT : A satisfying measure of how complicated it might be to find the roots of a polynomial is given by its Galois group. This edit partially addresses the concern that Wolfram was unable to factorize this polynomial as explicitly as I was, for example.
To provide an information explanation, the Galois group of a polynomial is a mathematical object that captures symmetries between roots. For example, consider the equation $x^3 - 1= 0$. The roots of this polynomial are given by $\omega , \omega^2$ and $\omega^3 = 1$ where $\omega$ is a complex cube root of unity. Notice the relations between the roots. This typically makes factoring easier.
A small Galois group (with respect to the biggest possible value of the size) typically indicates that a polynomial is "extremely simple". If you receive a high-degree polynomial which you're asked to factorize without a calculator (say, in an exam) then such a polynomial could have a very small Galois group (or is written very explicitly, typed wrongly etc. which makes the question easy for other reasons).
However, Galois groups are difficult to compute : and even fairly "obvious" symmetries such as reciprocity (also called Cohn or Gorenstein polynomials if some additional conditions are satisfied) are missed by engines that would much prefer numerical approximation to exact calculation if it meant a faster speed.
What follows is a vague explanation of how the Galois group is calculated in this case.
In this case, the most obvious symmetry is the relation that if $x$ is a root, so is $\frac 1x$. However, the reciprocal of the reciprocal returns you to the original number, so these relations are known as "involutions" because they are inverses of each other. There are three such involutions for three different roots, call them $r_1,r_2,r_3$.
Once one "forgets" about these involutions, then one is led to the polynomial $z^3-z^2+z-2$ that I derived earlier. This polynomial, as it turns out, doesn't have any polynomial relations among its roots. That has to do with a general result for cubic polynomials which is specified here.
From that thread, the Galois group of $z^3-z^2+z-2$ is of order $6$, the highest possible. In the case where the Galois group is of the highest possible order, the roots do not share any field relations with each other (so for example, even though $\bar{z_1}$ and $z_1$ are roots, it is not true that $\bar{z_1}$ is of the form $\frac{P(z_1)}{Q(z_1)}$ for $P,Q$ polynomials).
If these elements don't share any relations with each other, then you have $r_1,r_2,r_3$ which can be independently joined to any of these six permutations or not. That gives a choice of $6 \times 2^3 = \boxed{48}$ different permutations. The Galois group of $$
x^6-x^5+4x^4-4x^3+4x^2-x+1
$$
has size $48$ by making everything here rigorous.
The largest value that the Galois group could be? That is $6! = 720$.
Since $48$ is quite large but not very large, one is led to expect that there might be one or two simple relations between the roots, but not too many. This is reflected in what actually occurred.
Seeing that Galois group made me lick my lips. You've already seen why in the first part of this answer.
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.
Best Answer
The trick $x+\frac1x=t$ requires the polynomial to be a palindrome, so that is clearly not the case here. However that's not the only substitution possible to simply. Here, lets try something similar to what you have already done, persisting a bit more:
$ \begin{align} \frac{P(x)}{x^4} &= x^2+x+1-\frac2x-\frac1{x^2}+\frac1{x^4} \\ &= \left(x^2-\frac2x+\frac1{x^4}\right)+\left(x-\frac1{x^2} \right)+1\\ &= \left(x-\frac1{x^2} \right)^2+\left(x-\frac1{x^2} \right)+1\\ &=u^2+u+1 \tag{1} \end{align} $
where $u = x-\dfrac1{x^2} \tag{2}$
Now clearly $u$ is solvable as a quadratic from $(1)$, and for both values of $u$, the equation $(2)$ gives a cubic to solve for $x$. As both quadratics and cubics admit solutions in exact algebraic form (though using cumbersome radicals perhaps), we are done.