How to factor $x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1$ by hand?

The equation is palindromic (well, almost), so:

We can write it as $$x^3\left[x^3+x^2-3x+2+\frac 3x+\frac{1}{x^2}-\frac{1}{x^3}\right]$$ $$=x^3\left[(x^3-3x+\frac 3x-\frac{1}{x^3})+\left((x-\frac 1x)^2+2\right)+2\right]$$ $$=x^3\left[u^3+u^2+4\right],$$ where $u=x-\frac 1x$ And hence the factorization is $$x^3(u+2)(u^2-u+2)$$ which will give us the expected answer.

A possibility is to write a generic product

$x^6+x^5-3x^4+2x^3+3x^2+x-1=(x^4+a_3x^3+a_2x^2+a_1x+a_0)(x^2+b_1x+b_0)$

then expand the right-hand side and compare the two polynomials, obtaining

\begin{cases} &a_0 b_0=-1,\\ &a_0 b_1+a_1 b_0=1,\\ &a_0+a_1 b_1+a_2 b_0=3,\\ &a_1+a_2 b_1+a_3 b_0=2,\\ &a_2+a_3 b_1+b_0=-3,\\ &a_3+b_1=1 \end{cases}

that can be solved rather easily for integer solutions.
The same could be tried for a product of two third degree polynomials, without any (integer) solutions.

Note that $(x - 1/x)^{2k}$ gives $x^{2k} + x^{-2k}$ and other symmetric terms, and $(x - 1/x)^{2 k + 1} = x^{2 k + 1} - x^{- 2 k - 1}$ and other terms, it looks like due to the symmetry of the coefficients by dividing by $x^3$ you can reduce the degree to a cubic in $x - 1/x$, and go from there.