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.

If $x = a+b$, then $x^3 = (3ab)x + (a^3+b^3)$. 
(Conversely, if $x^3 - px - q = 0 $ then we may find $a,b$, such that $3ab = p, a^3+b^3=q$. 
using Cardano's method for solving the cubic equation.) 
Letting $w = e^{\frac {2i \pi}{3}} $ denote a cube root of unity, notice that the numbers $a+b,  wa + w^2b$, and $w^2a + wb$ all satisfy the same cubic equation, $x^3 = 3abx + (a^3+b^3)$, and in general the solutions to the equation 
$x^3 = 3w^iabx + (a^3+b^3) $ are given by $x = w^ja + w^{i-j}b$, for $j=0,1,2$. 
There are thus $9$ possible values of $(2+\sqrt{5})^{1/3} + (2-\sqrt{5})^{1/3}$, depending on which cube roots you take, and they each satisfy one of $3$ cubic equations.  But only one of these values, namely $1$, is real, let alone rational. Hope it helps. 
