<s>[INST] 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.
 [/INST] 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
$$
r^3=(a+a^{-1})^3=a^3+3a+3a^{-1}+a^{-3}=3r+52
$$
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