> 
> 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.