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.
 A: You are on the right track.
Let,
$$P(x)=ax^6-x^5+x^4+x^3-2x^2+1$$
We observe that, $0$ is not a possible root of $P(x)$. Therefore, we can divide all terms of the polynomial by $x^2\, (x\neq 0)\,:$
$$
\begin{align}\frac {P(x)}{x^2}&=ax^4-x^3+x+\left(x-\frac {1}{x}\right)^2\\
&=x^2\left(ax^2-\left(x-\frac 1x\right)\right)+\left(x-\frac {1}{x}\right)^2\\
&=ax^4-x^2\left(x-\frac 1x\right)+\left(x-\frac {1}{x}\right)^2\\
&=\color{red}{\left(x-\frac {1}{x}\right)^2}-\color{blue}{x^2}\color{red}{\left(x-\frac 1x\right)}+\color{blue}{ax^4}=0.
\end{align}$$
Finally, substitute $x-\frac {1}{x}=u$, then you have:
$$
\begin{align}
&u^2-ux^2+ax^4=0\\
\implies &\Delta_u=x^4(1-4a)\\
\implies &u_{1,2}=\frac {x^2\pm x^2\sqrt {1-4a}}{2}\\
\implies &u_{1,2}=x^2\left(\frac {1\pm \sqrt {1-4a}}{2}\right)\\
\implies &x-\frac {1}{x}=x^2\left(\frac {1\pm \sqrt {1-4a}}{2}\right)\\
\implies &\left(\frac {1\pm \sqrt {1-4a}}{2}\right)x^3-x^2+1=0.
\end{align}
$$
Note that, the last line is the result you want to achieve.
A: 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
$$\frac{1\pm\sqrt{1-4a}}{2}x^3-x^2+1=0$$
A: Define $t,u=\frac{1\pm\sqrt{1-4a}}{2a}$ as the roots of $ax^2-x+1$. Then the sextic splits into two cubics over $\mathbb Q(t)$:
$$ax^6-x^5+x^4+x^3-2x^2+1=a(x^3-tx^2+t)(x^3-ux^2+u)$$
This can be verified by re-expanding. Here the substitution is $a=\frac{t-1}{t^2}$ and not in $x$.
