Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification 
Solve the quartic polynomial :
$$x^4+x^3-2x+1=0$$
where $x\in\Bbb C$.
Algebraic, trigonometric and all possible methods are allowed.


I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.
I realized there is no any rational root, by the rational root theorem.
The harder part is, WolframAlpha says the factorisation over $\Bbb Q$ is impossible.
Another solution method can be considered as the quasi-symmetric equations approach. (divide by $x^2$).
$$x^2+\frac 1{x^2}+x-\frac 2x=0$$
But the substitution $z=x+\frac 1x$ doesn't make any sense.
I want to ask the question here to find possible smarter ways to solve the quartic.

You can easily observe that, the expression $(x-1)^2$ is almost included in the polynomial $P(x):=x^4+x^3-2x+1$.
Let's rewrite your polynomial as follows:
$$
\begin{align}P(x)=x^4+x^3-\color{red}{x^2}+\color{red}{x^2}-2x+1\end{align}
$$
Based on this observation, we will represent the polynomial $P(x)$ as a "quadratic" polynomial:
$$
\begin{align}P(x):&=x^4+x^3-x^2+(x-1)^2\\
&=x^2(x^2+x-1)+(x-1)^2\\
&=x^4+x^2(x-1)+(x-1)^2\\
&=\color{red}{(x-1)^2}+\color{blue}{x^2}\color{red}{(x-1)}+\color{blue}{x^4}\end{align}
$$
Letting $x-1=u$ and writing $P(x)=0$ leads to:
$$
\begin{align}&\color{red}{u^2}+\color{blue}{x^2}\color{red}{u}+\color{blue}{x^4}=0\\
&\Delta_{\color{red}{u}}=x^4-4x^4=-3x^4\\
\implies &u_{1,2}=\frac{-x^2\pm i\sqrt 3x^2}{2}\\
\implies &u_{1,2}=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\
\implies &x-1=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\
\implies &x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)-x+1=0.\end{align}
$$
Now, this is obvious that the given polynomial has no real roots.
Finally, you can complete the solution by using the quadratic formula to find all complex roots.
