How to show this equals 1 without "calculations" We have $$ \sqrt[3]{2 +\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} = 1 $$
Is there any way we can get this results through algebraic manipulations rather than just plugging it into a calculator?
Of course, $(2 +\sqrt{5}) + (2-\sqrt{5}) = 4 $, maybe this can in some way help?

I'll expand on Guillermo's comment: let
$$
a=\sqrt[3]{2+\sqrt{5}},\quad b=\sqrt[3]{2-\sqrt{5}}.
$$
You want to show that $a+b=1$. Consider
$$
(a+b)^3=a^3+3a^2b+3ab^3+b^3=(a^3+b^3)+3ab(a+b)=4+3\times(-1)(a+b)
$$
so if you let $x=(a+b)$ , then $x$ is a real number satisfying
$$
0=x^3+3x-4=x^3-x^2+x^2-x+4x-4\\
=(x-1)(x^2+x+4)=(x-1)\left[(x+0.5)^2+3.75\right].
$$
We infer that $x=1$.
