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?

There is an infinite family of solutions coming from the idea $(2+\sqrt 3)^3=26+15\sqrt 3$.  We can form $(2+\sqrt 3)^{3n}$ and find another solution.  The next one is $(2+\sqrt 3)^6=1351+780 \sqrt 3$ and $(1351+780\sqrt 3)^{(1/3)}+(1351-780\sqrt 3)^{(1/3)}=14$  There is a recurrence, if $(a,b)$ is a solution, the next is $(26a+45b,15a+26b)$ and so we get triplets $(1351,780,14),(70226,40545,52),(3650401,2107560,194),(189750626,109552575,724)
9863382151,5694626340,2702)$
and on.  I have not shown that these are all the solutions.
