<s>[INST] 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?
 [/INST] The solutions are of the form $\displaystyle(p, q)= \left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right)$, for any rational parameter $t$. To prove it, we start with $$\left(p+q\sqrt{r}\right)^{1/3}+\left(p-q\sqrt{r}\right)^{1/3}=n\tag{$\left(p,q,n,r\right)\in\mathbb{N}^{4}$}$$
and cube both sides using the identity $(a+b)^3=a^3+3ab(a+b)+b^3$ to, then, get $$\left(\frac{n^3-2p}{3n}\right)^3=p^2-rq^2,$$ which is a nicer form to work with. Keeping $n$ and $r$ fixed, we see that for every $p={1,2,3,\ldots}$ there is a solution $(p,q)$, where $\displaystyle q^2=\frac{1}{r}\left(p^2-\left(\frac{n^3-2p}{3n}\right)^3\right)$. When is this number a perfect square? Wolfram says it equals $$q^2 =\frac{(8p-n^3) (n^3+p)^2}{(3n)^2\cdot 3nr},$$ which reduces the question to when $\displaystyle \frac{8p-n^3}{3nr}$ is a perfect square, and you get solutions of the form $\displaystyle (p,q)=\left(p,\frac{n^3+p}{3n}\sqrt{\frac{8p-n^3}{3nr}}\right).$  Note that when $r=3$, this simplifies further to when $\displaystyle \frac{8p}{n}-n^2$ is a perfect square.

Now, we note that if $\displaystyle (p,q)=\left(p,\frac{n^3+p}{3n}\sqrt{\frac{8p-n^3}{3nr}}\right) \in\mathbb{Q}^2$, $\displaystyle\sqrt{\frac{8p-n^3}{3nr}}$ must be rational as well. Call this rational number $t$, our parameter. Then $8p=3t^2nr+n^3$. Substitute back to get $$(p,q)=\left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right).$$ This generates expressions like $$\left(\frac{2589437}{8}+\frac{56351}{4}\sqrt{11}\right)^{1/3}+\left(\frac{2589437}{8}-\frac{56351}{4}\sqrt{11}\right)^{1/3}=137$$
$$\left(\frac{11155}{4}+\frac{6069}{4}\sqrt{3}\right)^{1/3}+\left(\frac{11155}{4}-\frac{6069}{4}\sqrt{3}\right)^{1/3}=23$$
for whichever $r$ you want, the first using $(r,t,n)=(11,2,137)$ and the second $(r,t,n)=(3,7,23)$.
</s>