<s>[INST] How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$ find this equation roots:
$$2\sqrt[3]{2y-1}=y^3+1$$
My try: since
$$8(2y-1)=(y^3+1)^3=y^9+1+3y^3(y^3+1)$$
then
$$y^9+3y^6+3y^3-16y+9=0$$
Then I can't.Thank you someone can take hand  find the equation roots.
 [/INST] Let $f(y) = \frac12 (y^3+1)$, we have
$$2\sqrt[3]{2y-1} = y^3+1\quad\iff\quad f^{-1}(y) = f(y) \quad\implies\quad y = f(f(y))$$
Since $f(y)$ is a strictly increasing function in $y$, we have


*

*If $f(x) > x$, then $f(f(x)) > x$.

*If $f(x) < x$, then $f(f(x)) < x$.


This means 
$$\begin{align}
y = f(f(y)) \implies & y = f(y)\\
\iff & y^3+1 - 2y = 0\\
\iff & (y^2 + y -1 )(y-1) = 0\\
\implies & y = 1\text{ or }  \frac{-1\pm \sqrt{5}}{2}
\end{align}$$
</s>