<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] 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$.
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