Reducing $ax^6-x^5+x^4+x^3-2x^2+1=0$ to a cubic equation using algebraic substitutions 
Use algebraic substitutions and reduce the sextic equation to the cubic equation, where $a$ is a real number:
$$ax^6-x^5+x^4+x^3-2x^2+1=0$$

My attempts.
First, I tried to use the Rational root theorem, when $a$ is an integer $x=\pm 1$, but this implies $a=0$ and this is not always correct. Then I realized that,
$$x^4-2x^2+1=(x^2-1)^2$$
is a perfect square. So, I tried to write the original equation as
$$ax^4-x^3+x+\bigg(x-\frac 1x\bigg)^2=0$$
$$x^2\bigg(ax^2-x+\frac 1x\bigg)+\bigg(x-\frac 1x\bigg)^2=0$$
But I failed again. I couldn't spot the palindromic property.

The coefficient $a$ should be something difficult to deal with, so isolating $a$ might help.
We know that $x=0$ is not a root, so we can divide by $x^6$ to have
$$a=\frac{x^5-x^4-x^3+2x^2-1}{x^6}$$
$$a=\frac 1x-\frac{1}{x^2}-\frac{1}{x^3}+\frac{\color{red}2}{x^4}-\frac{1}{x^6}$$
Now, we can focus on the RHS which has no $a$, and we are happy if the RHS is of the form $uX^2+vX+w$. The coefficient $\color{red}2$ reminds me of $A^2+\color{red}2AB+B^2=(A+B)^2$. So, we may have $\bigg(\dfrac 1x\pm \dfrac{1}{x^3}\bigg)^2$, then we note that
$$a=\bigg(\frac 1x-\frac{1}{x^3}\bigg)-\bigg(\frac 1x-\frac{1}{x^3}\bigg)^2$$
Letting $X=\dfrac 1x-\dfrac{1}{x^3}$, we have
$$X^2-X+a=0\implies \frac 1x-\frac{1}{x^3}=X=\frac{1\pm\sqrt{1-4a}}{2}$$
and finally get
$$\frac{1\pm\sqrt{1-4a}}{2}x^3-x^2+1=0$$
