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.

Use the general formula $x^3+y^3=(x+y)\left((x+y)^2-3xy\right)$.
Here $x=\sqrt[3]{2+\sqrt{5}}$ and $y=\sqrt[3]{2-\sqrt{5}}$.
Let $a=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ and
$$b=\sqrt[3]{2+\sqrt{5}}\sqrt[3]{2-\sqrt{5}}=\sqrt[3]{2^2-(\sqrt{5})^2}=-1$$
$$(2+\sqrt{5})+(2-\sqrt{5})=4$$
$$=a\left(a^2-3b\right)=a^3+3a$$
$$a^3+3a-4=0$$
You can apply Rational Root Theorem/Test, polynomial division, fundamental theorem of algebra, etc.
$$(a-1)\left(a^2+a+4\right)=0$$
$a^2+a+4=0$ has no real solutions, but clearly $a$ is real, so $a=1$. 
