Solving the equation $\frac{x^7}{7}=1+10^{1/7}x(x^2-10^{1/7})^2$

Question

$$\frac{x^7}{7}=1+10^{1/7}x(x^2-10^{1/7})^2$$ Find $x$ where $x$ is real.

Answer 1

This is really a trick question. Let $\alpha = 10^{1/14}$, the equation we have can be rewritten as $$P(x) = 0 \quad\text{ where }\quad P(x) = \frac{x^7}{7} - 1 - \alpha^2 x (x^2-\alpha^2)^2$$ Let $x = \alpha y$, we can further simplify the equation

$$P(x) = 0 \quad\iff\quad\alpha^{-7} P(\alpha y) = 0 \quad\iff\quad\frac{y^7}{7} - y^5 + 2y^3 - y - \alpha^{-7} = 0$$ The polynomial in $y$ on the RHS looks familiar. In fact, aside from the constant term, it is proportional to a Chebyshev polynomial of the first kind to degree 7:

$$\frac{y^7}{7} - y^5 + 2y^3 - y = \frac{2}{7}T_7\left(\frac{y}{2}\right) = \begin{cases} \frac{2}{7}\cos\left(7\cos^{-1}\left(\frac{y}{2}\right)\right), & |y| \le 2\\ \frac{2}{7}\cosh\left(7\cosh^{-1}\left(\frac{y}{2}\right)\right),& |y| \ge 2 \end{cases}$$

Since $\alpha^{7} = \sqrt{10}$, we get $$\begin{align} & y = 2\cosh\left(\frac17\cosh^{-1}\left(\frac{7}{2\sqrt{10}}\right)\right)\\ \iff & x = 2\times 10^{1/14}\cosh\left(\frac17\cosh^{-1}\left(\frac{7}{2\sqrt{10}}\right)\right) \sim 2.362588464315639 \end{align}$$ Update

I'm missing an obvious simplification. Let $\theta = \cosh^{-1}\left(\frac{7}{2\sqrt{10}}\right)$, we have

$$ e^\theta = \frac{7 + \sqrt{7^2 - (2\sqrt{10})^2}}{2\sqrt{10}} = \frac{5}{\sqrt{10}} \;\;\implies\;\;x = 10^{1/14} \left(e^{\frac{\theta}{7}} + e^{-\frac{\theta}{7}}\right) = 5^{1/7} + 2^{1/7} $$