<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] An easy way is to simplify the nested radical, and then combine like terms.
An easy way to denest $\sqrt[m]{A+B\sqrt[n]{C}}$ is to assume the form $a+b\sqrt[n]{C}$ and expand both sides with the binomial theorem. $\sqrt[3]{2+\sqrt5}$ becomes$$2+\sqrt5=(a^3+15ab^2)+(3a^2b+5b^3)\sqrt5\tag1$$
From which we get a system of equations\begin{align*} & a^3+15ab^2=2\tag2\\ & 3a^2b+5b^3=1\tag3\end{align*}
Cross multiplying, we get$$a^3+15ab^2=6a^2b+10b^3\\a^3-6a^2b+15ab^2-10b^3=0\tag4$$
Dividing the last equation of $(4)$ by $b^3$ and substituting $x=a/b$ gives the cubic$$x^3-6x^2+15x-10=0\implies x=1$$
Therefore, $a=b$ and plugging that back into the system, we get$$a^3+15ab^2=2\implies 16b^3=2\implies a=b=\dfrac 12$$
Hence, $\sqrt[3]{2+\sqrt5}=\dfrac 12+\dfrac {\sqrt5} 2$ and similarly, $\sqrt[3]{2-\sqrt5}=\dfrac 12-\dfrac {\sqrt5}2$.
Therefore, we have$$\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}=\dfrac 12+\dfrac 12+\dfrac {\sqrt5}2-\dfrac {\sqrt5}2=\boxed{1}\tag5$$
</s>