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?
 A: Using $\bullet\; $ If $a+b+c=0\;,$ Then $a^3+b^3+c^3=3abc$
Here $$(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}+(-x) = 0$$
Then $$a+\sqrt{a^2+b^3}+a-\sqrt{a^2+b^3}-x^3=3\left[\left(a+\sqrt{a^2+b^3}\right)^{\frac{1}{3}}\cdot \left(a-\sqrt{a^2+b^3}\right)^{\frac{1}{3}}\right]\cdot -x$$
So $$2a-x^3=-3x\cdot -b\Rightarrow x^3+3bx-2a=0$$
A: Let $p = (a+\sqrt {a^2+b^3})^\frac {1}{3}$ and $q = (a-\sqrt {a^2+b^3})^\frac {1}{3}$
Hence,
$p + q = x$ -- (I)
$p^3 + q^3 = (a+\sqrt {a^2+b^3}) + (a-\sqrt {a^2+b^3}) = 2a$ --(II)
$pq = (a^2 - (a^2+b^3))^\frac{1}{3} = (-b^3)^\frac{1}{3} = -b$ --(III)
Now,
$(p+q)^3 = p^3 + q^3 + 3pq(p+q)$
Using the statements I, II and III in the above expansion, we get,
$$x^3 = 2a + 3(-b)x$$
Re-arranging the terms,
$$x^3 + 3bx = 2a$$
as desired.
A: Hint: $$(x+y)^3=x^3+y^3+3x^2y+3xy^2=x^3+y^3+3xy(x+y)$$
A: 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.
A: 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}
