Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$. Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the family are drawn on the same picture, one gets (click to enlarge):

What I'm calling "pillars" make their appearance. They are located on precise abscissas $x_{c/d}$ which are algebraic numbers, roots of $X^c-(1-X)^d$ polynomials for some positive rational $c/d$ (in irreducible form). To each pillar is associated a vertical wavelength (clearly visible on the picture) which is equal to $-\frac{\ln(x_{c/d})}{d}=-\frac{\ln(1-x_{c/d})}{c}$. I compared the wavelength of the $(c/d=5)$ pillar (for which $x_5$ is the multiplicative inverse of $\rho$, the plastic number) with the wavelength of the $(c/d=3/2)$ pillar (marked green), and found numerically that they were equal. So, the asked question is to prove that, in any of its forms:

*

*$$ -\frac{\ln(x_{3/2})}{2} = -\frac{\ln(x_{5})}{1} $$

*$$ x_{3/2} = x_{5}^2 $$

*the real root of $X^3-(1-X)^2$ is the square of the real root of $X^5-(1-X)^1$

*the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$
 A: Let $\alpha$ denote the real root of $X^3+X^2-1$. Then $\alpha^3 = 1-\alpha^2$. If we plug in $\alpha^2$ into $X^3-X^2+2X-1$, we obtain
\begin{align*}
   (\alpha^2)^3-(\alpha^2)^2+2(\alpha^2)-1 &= (1-\alpha^2)^2-\alpha^4+2\alpha^2-1\\
   &=(1-\alpha^2)^2-(1-\alpha^2)^2=0,
\end{align*}
so $\alpha^2$ is a root of $X^3-X^2+2X-1$.
A: Let $f(x)=x^3-x^2+2x-1, g(x)=x^3+x^2-1$. We have
$$-g(x)\cdot g(-x)=-(x^3+x^2-1)((-x)^3+(-x)^2-1)\\
=(x^3+x^2-1)(x^3-x^2+1) = x^6 - (x^2-1)^2\\
=x^6-x^4+2x^2-1=f(x^2)$$
Now if $g(x)=(x-r)(x-s)(x-t)$, then
$$f(x^2)= - (x-r)(x-s)(x-t) (-x-r)(-x-s)(-x-t)\\=(x^2-r^2)(x^2-s^2)(x^2-t^2)
\\\implies f(x)=(x-r^2)(x-s^2)(x-t^2)$$
and we are done.
A: We will prove the last statement.
Let $\alpha$ be a root of $p_1:= X^3 + X^2-1$, i.e., $\alpha^3 +\alpha^2-1 = 0$. We will prove that $\beta := \alpha^2$ is a root of $p_2:= X^3-X^2+2X-1$.
We have the following identities:
$$\beta = \alpha^2,$$
$$\beta^2 = \alpha^4 = \alpha\cdot \alpha^3 = \alpha (-\alpha^2 +1) = -\alpha^3 + \alpha,$$
$$\beta^3 = \alpha^6 = (\alpha^3)^2 = (-\alpha^2+1)^2 = \alpha^4 -2\alpha^2 +1 = -\alpha^3 -2\alpha^2 + \alpha +1.$$
Then,
$$\beta^3 - \beta^2 +2\beta -1 = (-\alpha^3 -2\alpha^2 + \alpha +1) - (-\alpha^3 + \alpha) +2(\alpha^2) -1 = 0.$$
Thus $\beta$ is a root of $p_2$. Note that we did not require $\alpha$ to be real, this means that the roots of $p_2$ are precisely the squares of the roots of $p_1$. Particularly, the square of the real root of $p_1$ is a (the) real root of $p_2$.
