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

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 $\cos(\theta)=\frac{R}{\sqrt{-Q^3}}$
