<s>[INST] $\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ 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
 [/INST] By the rational root theorem, a solution of $$x^3-x-n=0$$ must be a divisor of $-n.$  Since $n\leq50$ such a solution can't be very large. By definition of $x$ we have $x\geq0$.  If $x=4,$ then $x^3 =64$ and there's already no hope.  So $x$ must be one of $0, 1, 2, 3.$  
$$\begin{align}
x&=0\implies n=0\\
x&=1\implies n=0\\
x&=2\implies n=6\\
x&=3\implies n=24
\end{align}$$
Now you have to test these combinations of $n$ and $x$ to see $n+x$ is a perfect square.  The only ones that work are the ones where $n=0$ and obviously $x=1$ is wrong in that case.  So we are left with $n=0,$ which isn't very interesting.
I suspect that there really is supposed to be a cube root on the outside, and the answer is $n=24, x=3.$   
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