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.
 A: Another approach:
Put $x=\sqrt[3]{2y-1}$ then $x^3=2y-1$. So, $x^3+1=2y$
Then we have the system of equation:
$\left\{\begin{matrix} x^3+1=2y\\y^3+1=2x \end{matrix}\right.$
Subtracting these two equations from each other, we have
$(x-y)(x^2+xy+y^2+2)=0$
which gives $x=y$ as ($x^2+xy+y^2+2>0 $  $\forall$ $x,y$)
Hence we have $y=\sqrt[3]{2y-1}$ 
$\Rightarrow y^3-2y+1=0$ 
$\Leftrightarrow(y-1)(y^2+y-1)=0$
Therefore, $y=1$ or $y=\frac{-1\pm \sqrt{5} }{2}$
A: $$y^9+3y^6+3y^3-16y+9=0\iff(y-1)(y^2 + y - 1)(y^6 + 2y^4 + 2y^3 + 4y^2 + 2y +9) = 0$$
Three real-valued roots: $$y = 1, y = \dfrac{-1 \pm \sqrt 5}{2}$$
I noticed by inspection that $y = 1$ solves the equation. Then, using polynomial division, and manipulation with the resulting quotient, was able to find the quadratic factor. The sixth degree polynomial has no real roots.
Additional tip: Sometimes it helps to graph the equation to learn how many real roots there are: look for points intersecting the x-axis. Here's the graph of your 9th-degree polynomial:

A: 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 
$$\begin{align}
y = f(f(y)) \implies & y = f(y)\\
\iff & y^3+1 - 2y = 0\\
\iff & (y^2 + y -1 )(y-1) = 0\\
\implies & y = 1\text{ or }  \frac{-1\pm \sqrt{5}}{2}
\end{align}$$
