<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] Let $x=(2+\sqrt 5)^{1/3}$ and $y=(2-\sqrt 5)^{1/3} .$
We have $5>x^3>4$ and $0>y^3>-1$ so $2>x>1$ and $0>y>-1,$ so $2>x+y>0.$ Therefore $$x+y\in \mathbb Z \iff x+y=1.$$ We have $xy= (x^3y^3)^{1/3}=(-1)^{1/3}=-1$ and $x^3+y^3=4.$ So we have  $$(x+y)^3=x^3+y^3+3xy(x+y)=4-3(x+y).$$ So $x+y$ is a positive solution to the equation $z^3+3z=4.$  The function $f(z)=z^3+3z$ is strictly increasing for $z>0$ (As it is the sum of the strictly increasing functions $z^3$ and $3z$.)
So $f(z)=4$ has at most one positive solution, which, if it exists, is  equal to $x+y.$ Therefore $$x+y\in \mathbb Z \iff 1^3+3\cdot 1=4.$$
For insight on why this kind of thing happens, see Cardano's Method for solving cubic equations. 
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