<s>[INST] Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
 [/INST] Take the cube of the expression and expand it. You get
\begin{multline}
\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)^3
\\= 2 + \sqrt{5} + 2 - \sqrt{5} + 3\sqrt[3]{(2 + \sqrt{5})(2 - \sqrt{5})}\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)
\\ = 4 - 3 \left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)
\end{multline}
Thus, $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ solves $x^3 = 4 - 3x$. It's not too difficult to find that $1$ is a solution to this equation.
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