If $x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$ If $$x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$$ then prove that $x^3+3bx=2a$.
By observing the given question, I thought about cubing on both sides. But it becomes quiet vague and complex. 
Can anyone help me with a simpler proof?

The cube of a binomial is
$$
(u+v)^3=u^3+v^3+3uv(u+v)
$$
Let us observe that in this case
\begin{align}
&u^3+v^3&&=2a\\
&uv&&=[a^2-(a^2+b^3)]^{1/3}=-b\\
&u+v&&=x
\end{align}
