Solve for x (radicals)

Question

Solve this equation for x

$ \sqrt[3]{\sqrt{x}+3} + \sqrt[3]{13-\sqrt{x}} = 4 $

I tried to put a condition for x to be $ \ge 0 $ but then do I need to cube the expressions or what?

Answer 1

HINT:

$$4^3=3+13+3\cdot4\cdot\sqrt[3]{39+10\sqrt x-x}$$