Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this result can be generalized to other triples of
natural numbers. 

Question. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad  \text{where }\left( p,q,n\right)
\in  \mathbb{N} ^{3}.\tag{2} \end{equation} $$

For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$ 
$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) 
\sqrt{3}
$$
and solve the system
$$
\left\{ 
\begin{array}{c}
a^{3}+9ab^{2}=26 \\ 
a^{2}b+b^{3}=5.
\end{array}
\right. 
$$
A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.
For $(2)$ the very same idea yields
$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$
and
$$
\left\{ 
\begin{array}{c}
a^{3}+9ab^{2}=p \\ 
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$
I tried to solve this system for $a,b$ but since the solution is of the form
$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$
where $x$ satisfies the cubic equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach. 

Is this problem solvable, at least partially?

Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?

Yikes. I am hopind I didn't do any mistake, but that would be a miracle :)
Let $x= \left( p+q\sqrt{3}\right) ^{1/3}\,;\, y= \left(
p-q\sqrt{3}\right) ^{1/3}$.
Then 
$$xy= (p^2-3q^2)^\frac{1}{3} $$
Hence
$$2p=x^3+y^3=(x+y)^3-3xy(x+y)=n^3-3n\sqrt[3]{p^2-3q^2}$$
This shows that $\sqrt[3]{p^2-3q^2}$ must be rational, hence integer.
Let $p^2-3q^2=k^3 (*)$. Then $y=\frac{k}{x}$ and thus
$$x+\frac{k}{x} =n \Rightarrow x^2-nx+k=0 \Rightarrow x= \frac{n \pm \sqrt{n^2-4k}}{2}  \,.$$
The two roots of this equation must be $x$ and $y$, and thus we get:
$$x=\frac{n + \sqrt{n^2-4k}}{2} \,;\, y= \frac{n - \sqrt{n^2-4k}}{2} \,.$$
Then
$$p+q\sqrt{3}=\left(\frac{n + \sqrt{n^2-4k}}{2} \right)^3 \,.$$
From here, we get that $\sqrt{n^2-4k} \notin \mathbb Q$ and hence
$$p=\frac{n^3+3n^3-12nk}{8}=\frac{n^3-3nk}{2}$$
$$q\sqrt{3}=\frac{3n^2+n^2-4k}{8}\sqrt{n^2-4k}=\frac{n^2-k}{2}\sqrt{n^2-4k}$$
Thus, $n^2-4k=3u^2 $, for some $u$, hence
$$k=\frac{n^2-3u^2}{4} (***)$$
Thus,  we get:
$$p=\frac{n^3+9nu^2}{8}$$
$$q=\frac{3n^2+3u^2}{8}u=\frac{3n^2u+3u^3}{8}$$
With this choice we have
$$p+q\sqrt{3}=(\frac{n+u\sqrt{3}}{2})^3$$
$$p-q\sqrt{3}=(\frac{n-u\sqrt{3}}{2})^3$$
Note that  $p,q$ integers if and only if $8|n(n^2+u^2)$ and $8|u(n^2+u^2)$. It is easy to check that the first one cannot happen if $n$ is odd. Thus, $n$ must be even and $ 8|u(n^2+u^2) \Rightarrow 2|u^3 \Rightarrow u$ even.
If $n=2s, u=2t$ we get the general solution
$$p=s^3+9st^2$$
$$q=3s^2t+3t^3$$
$$n=2s$$
Note that in this case, 
$$\sqrt[3]{p+q\sqrt{3} }=s+t\sqrt{3}$$
$$\sqrt[3]{p-q\sqrt{3} }=s-t\sqrt{3}$$
