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?

Notice that
$$  \left( \sqrt[3]{a+b}+\sqrt[3]{a-b} \right)^3= 2a + 3\cdot \sqrt[3]{a^2-b^2}\left(\sqrt[3]{a-b}+\sqrt[3]{a+b} \right)$$
Now with your question
Consider
$$x= \sqrt[3]{a+\sqrt {a^2+b^3}}  +\sqrt[3]{a-\sqrt {a^2+b^3}} $$
Let $\sqrt{a^2+b^3} = c$
$$ x= \sqrt[3]{a+c}  +\sqrt[3]{a-c} $$
$$ \Leftrightarrow x^3 = 2a + 3 \cdot \sqrt[3]{a^2-c^2}\left(\sqrt[3]{a-c}+\sqrt[3]{a+c} \right)  $$
$$ x^3 = 2a + 3\cdot \sqrt[3]{a^2-c^2}(x)$$
I think you can take it from here.
