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?

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.
