How to show that $\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} \in \mathbb{Z}$? How to show that the following is true? $$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} \in \mathbb{Z}$$
I have tried to set $$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} = r,$$ $$a=\sqrt[3]{26+15\sqrt{3}},$$ $$b=\sqrt[3]{26-15\sqrt{3}},$$ and used the identity $$(a^{1/3} + b^{1/3})^3 = a + b + 3(ab)^{1/3}(a^{1/3} + b^{1/3})$$ but I got nowhere. I am stuck at $$\left(a^{1/3}+b^{1/3}\right)^3=52+ 3 \cdot \left(a^{1/3}+b^{1/3}\right)$$ 
I'd be glad at any help.
 A: Find $p$ and $q$ such that
$$
-\frac{q}{2}=26
\qquad
\frac{q^2}{4}+\frac{p^3}{27}=675
$$
We get $q=-52$ and $p=-3$. So your number is a root of
$$
x^3-3x-52=0
$$
and you can see that $4$ is the only real solution. Cardan's formula tells you that
$$
\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}=4
$$
Alternatively,
$$
\frac{1}{26-15\sqrt{3}}=\frac{26+15\sqrt{3}}{26^2-675}=26+15\sqrt{3}
$$
so $b=a^{-1}$. Then
$$
r^3=(a+a^{-1})^3=a^3+3a+3a^{-1}+a^{-3}=3r+52
$$
A: Set $ r=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}$. Then $$r^3= 26+15\sqrt{3}+3\sqrt[3]{( 26+15\sqrt{3})^2}\sqrt[3]{ 26-15\sqrt{3}}+3\sqrt[3]{ 26+15\sqrt{3}}\sqrt[3]{( 26-15\sqrt{3})^2}+ 26-15\sqrt{3}=52+3( \sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{ 26-15\sqrt{3}}\;)=52+3r.$$ Therefore $r^3-3r-52=0$. A quick factorization gives us $(r-4)(r^2+4r+13)=0$. Since $ r^2+4r+13$ doesn't have real roots, we deduce that $ r=4$.
A: We have
$$
26 \pm 15\sqrt{3} = (2 \pm \sqrt{3})^3
$$
(by educated guessing, see note below) which gives us
$$
\sqrt[3]{26 + 15\sqrt{3}} + \sqrt[3]{26-15\sqrt3} = 2+\sqrt3+2-\sqrt3 = 4
$$

Note on guessing: I guessed that $26 + 15\sqrt3$ would be a nice cube of the form $(n + m\sqrt3)^3$ for integers (or at least rational numbers) $n, m$. If that were indeed the case, then we would have
$$
26 + 15\sqrt3 = (n + m\sqrt3)^3 = n^3 + 3n^2m\sqrt3 + 9nm^2 + 3m^3\sqrt3
$$
which then becomes
$$
n^3 + 9nm^2 = 26 \qquad\bigwedge\qquad 3(n^2m + m^3) = 15
$$
The second equation becomes $(n^2 + m^2)m = 5$ which has $n = 2, m = 1$ as the nicest solution. It turns out that this solution also solves the first equation, so we're done.
