$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ 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
 A: $n = 0$ 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.
A: If zero is a natural number, then $n=0$
Otherwise, wolfram alpha gives integer solutions of $x$ for $n=6$ and $n=24$
$$x^3-x-24=0 \implies x = 3$$
$$x^3-x-6=0 \implies x = 2$$
There's probably some algebra or number theory way to go about ensuring that $x$ is an integer if $n$ is an integer.
Update: it's the rational root theorem.
A: 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 $n=24, x=3.$   
A: Using the formula for finding the roots of the general cubic equation we obtain the numerical criterion:
$$
\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+}\ldots}}}\in \mathbb{N} \Longleftrightarrow
 \sqrt[3]{\dfrac{n}{2}+\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}+\sqrt[3]{\dfrac{n}{2}-\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}
\in \mathbb{N}
$$
Let  the cubic equation $a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$. In this case we have 
$
a=1\qquad b=0 \qquad c=-1 \quad \mbox{ and } \quad d=-n
$
and 
$
1\cdot x^3-0\cdot x^2-1\cdot x-n=0. Let
$
$Q = \dfrac {3 a c - b^2} {9 a^2}=-\dfrac{1}{3}$
$R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}=\dfrac{n}{2}$
$S = \sqrt [3] {R + \sqrt{Q^3 + R^2}}=\sqrt[3]{\dfrac{n}{2}+\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}$
$T = \sqrt [3] {R - \sqrt{Q^3 + R^2}}=\sqrt[3]{\dfrac{n}{2}-\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}$
If  $D = Q^3 + R^2\geq 0 $ ( In this case $D=\dfrac{n^2}{4}-\dfrac{1}{27} >0$ ) then the cubic equation has solutions:
$x_1 = S + T - \dfrac b {3 a}= \sqrt[3]{\dfrac{n}{2}+\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}+\sqrt[3]{\dfrac{n}{2}-\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}$
$x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \left({S - T}\right)$
$x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \left({S - T}\right)$
If  $D = Q^3 + R^2<0 $ then the cubic equation has solutions:
$x_1 = 2 \sqrt {-Q} \cos \left({\dfrac \theta 3}\right) - \dfrac b {3 a}$
$x_2 = 2 \sqrt {-Q} \cos \left({\dfrac \theta 3 + \dfrac {2 \pi} 3}\right) - \dfrac b {3 a}$
$x_3 = 2 \sqrt {-Q} \cos \left({\dfrac \theta 3 + \dfrac {4 \pi} 3}\right) - \dfrac b {3 a}$
where $\cos(\theta)=\frac{R}{\sqrt{-Q^3}}$
A: 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 $x$   
