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

Reshetnikov proved that

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?

For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$

and solve the system

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

and

I tried to solve this system for $a,b$ but since the solution is of the form

where $x$ satisfies the cubic equation

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?

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
and cube both sides using the identity $(a+b)^3=a^3+3ab(a+b)+b^3$ to, then, get 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 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 This generates expressions like
for whichever $r$ you want, the first using $(r,t,n)=(11,2,137)$ and the second $(r,t,n)=(3,7,23)$.