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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.

user1094359
  • 652
  • I have a thought: Try writing $ax^6-x^5+x^4+x^3-2x^2+1$ as $ a(x^2+bx+c)^3 + d(x^2+bx+c)^2 + e(x^2+bx+c) + f = 0.$ If you can find the values of $b,c,d,e,f$ then I think that more or less answers the question? – Adam Rubinson Sep 24 '22 at 21:30
  • You are looking for numbers $a_2,a_1,a_0$, $b_2,b_1b_$, and $c_0$ such that $(a_2x^2+a_1x+a_0)^3 +(b_2x^2+b_1x+b_0) +c_0 = ax^6-x5+x^4 +x^3-2x^2 +1$. Expand the left size and compare coefficients. – Salcio Sep 24 '22 at 21:33

3 Answers3

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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 $x$.

Parcly Taxel
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4

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.

lone student
  • 14,709
2

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 $$\frac{1\pm\sqrt{1-4a}}{2}x^3-x^2+1=0$$

mathlove
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