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.
 A: You correctly set $x^3=2y-1$, so the equation becomes
$$
\begin{cases}
2y=x^3+1\\
2x=y^3+1
\end{cases}
$$
Subtract them:
$$
2(y-x)=x^3-y^3
$$
or
$$
(x-y)(x^2+xy+y^2)+2(x-y)=0
$$
This factors as
$$
(x-y)(x^2+xy+y^2+2)=0
$$
but the second factor is positive for all $x$ and $y$, because $x^2+xy+y^2\ge0$.
Therefore you conclude $x=y$, so plugging in the equation $2y=x^3+1$ you get
$$
x^3-2x+1=0
$$
that's easy to solve because it factors as $(x-1)(x^2+x-1)=0$.

There is a different way to solve the particular problem, but using more general principles (thanks to ivancho for having suggested it).
Your equation is in the form
$$
f^{-1}(y)=f(y)
$$
where
$$
f(y)=\frac{y^3+1}{2}
$$
is an increasing function (hence invertible) taking on every real value. Decreasing would be the same, of course.
If $y$ is a solution of your equation and we set $x=f(y)$, then from $f^{-1}(y)=f(y)$ we get $f^{-1}(y)=x$ and so $y=f(x)$.
Suppose $(x,y)$ is a solution of $x=f(y)$ and $y=f(x)$. If we had $x<y$, then, by the fact that $f$ is increasing, we would have
$$
f(x)<f(y)
$$
so $y<x$. Similarly, from $x>y$ we get $f(x)>f(y)$ or $y>x$. In both cases we get a contradiction.
A: 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.
A: Letting $t=\sqrt[3]{2y-1}$ we have $y=\dfrac{t^3+1}2$ . Replacing, we have a nonic equation in t, which has $t=1$ and $t=\dfrac{-1\pm\sqrt5}2$ among its roots, the remaining six belonging to an irreducible sextic, all of which are complex.
