<s>[INST] 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?
 [/INST] 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 $$2a-x^3=-3x\cdot -b\Rightarrow x^3+3bx-2a=0$$
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