Solving the equation $\frac{x^7}{7}=1+10^{1/7}x(x^2-10^{1/7})^2$ $$\frac{x^7}{7}=1+10^{1/7}x(x^2-10^{1/7})^2$$ Find $x$ where $x$ is real.
 A: 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
$$
e^\theta = \frac{7 + \sqrt{7^2 - (2\sqrt{10})^2}}{2\sqrt{10}} = \frac{5}{\sqrt{10}}
\;\;\implies\;\;x = 10^{1/14} \left(e^{\frac{\theta}{7}} + e^{-\frac{\theta}{7}}\right) = 5^{1/7} + 2^{1/7}
$$
