how to solve $2(2y-1)^{\frac{1}{3}}=y^3+1$ how can I solve $2(2y-1)^{\frac{1}{3}}=y^3+1$
I got by $x=\frac{y^3+1}{2}$ that $y=\frac{x^3+1}{2}$ but I was told earlier that can't say that $x=y$.
So I can I solve this equation?
Thanks.

By cubing you get
$$ y^9+3 y^6+3 y^3 - 16 y + 9 = 0$$
It factors (using Wolfram alpha) into
$$ (y-1) (y^2+y-1) (y^6+2 y^4+2 y^3+4 y^2+2 y+9) = 0$$
According to WA, all the roots of the last polynomial are complex. The second one has $ \frac{(-1-\sqrt{5})}{2}$ and $ \frac{(\sqrt{5}-1)}{2}$ as roots. I don't know if you can get exact forms for the other (complex) roots.
