$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is 
The number of natural number $n\leq 50$ such that
$\displaystyle \sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is

Try: Let $\displaystyle x=\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}$
So $\displaystyle x=\sqrt[3]{n+x}\Rightarrow x^3=n+x\Rightarrow x^3-x-n=0$
could some help me how to solve it, Thanks

With $$\displaystyle \sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$$ We get $$x^3-x-n=0$$
Thus $$n=x^3-x=x(x-1)(x+1)$$
Therefore for any natural number $x>1$ we have a natural number $n$ satisfying $$n=x^3-x=x(x-1)(x+1)$$
For example with $x=5$ we get $n=120$ and for $x=10$ we get $n=990.$
Another way of saying that is for any three natural numbers, we can have their product as $n$ and the middle of the three natural numbers to be $x$   
