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.

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 $y=\frac{-1\pm \sqrt{5} }{2}$
