How to show that$\ \sqrt[3]{ \sqrt{y^2-x}+y}+ \sqrt[3]{-\sqrt{y^2-x}+y} = k \implies y = \frac{k\left(k^2-3 \sqrt[3]{x}\right)}{2}$?

Question

We also have $\ x \ne y $, $\ y > 1$, $\ 0<x<1$,$\ k \ne 0$.
I have tried on my own, by canceling out the roots, but they keep on appearing. I guess that is not the right way. Thanks in advance.

Answer 1

$$k^3= \sqrt{y^2-x}+y+ (-\sqrt{y^2-x}+y)+3\sqrt[3]{(-\sqrt{y^2-x}+y)(\sqrt{y^2-x}+y)}(k)$$

As $(-\sqrt{y^2-x}+y)(\sqrt{y^2-x}+y)=y^2-(y^2-x)=x,$

this reduces to $k^3=2y+3k\sqrt[3]x$

Hope the destination is easily visible from here