What is the form of the roots for a cubic equation of only the cubic variable?

Question

For a quadratic equation $y^2=x+b$ we can say that the solutions for $y$ are $$y_1=\sqrt{x+b}$$ $$y_2=-\sqrt{x+b}$$ For any cubic equation $y^3=x+b$, what is the form of its roots is going to be?

Answer 1

Nominally… $$y=\sqrt[3]{x+b}$$ But there is a caveat. There are three solutions; one is as above, over the reals, but the other two are complex. Let $\omega$ satisfy $\omega^3=1$ but $\omega\ne1$ (thus $\omega=-\frac12\pm\frac{\sqrt3}2i$). Then all the solutions for $y$ are $$\sqrt[3]{x+b},\ \omega\sqrt[3]{x+b},\ \omega^2\sqrt[3]{x+b}$$ where the cube root is taken to be real, as presented above.