$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ 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

$n = 0$ would fit. Now let's assume $n > 0$.
I think you are on the right track. If the given expression, call it $y$, is a natural number, then $x = y^2 - n$ must also be an integer - and it is strictly positive, so it must be a natural number.
Then from $x^3-x-n = 0$ and $x$ being an integer, we conclude that $x$ must be a divisor of $n$, and in particular $x \le n$.
Since $n \le 50$, we get $x^3 = x + n \le 100$, which means $x \le 4$. Try $x = 1, 2, 3, 4$, compute the respective $n$, and see if $y$ is still an integer.
