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?
 A: $u^3=2+\sqrt 5$, $v^3=2-\sqrt 5$
$u^3v^3=4-5=-1$ so $uv=-1$.
$u^3+v^3=4$
$(u+v)^3=u^3+v^3+3uv(u+v)=4-3(u+v)$.
That is, $u+v$ is a root of the equation $X^3+3X-4=0$ (1). The derivative of this polynomial is $3X^2+3$ which has no roots. By Rolle's theorem, that means that (1) has only one real root. You can easily check that $1$ is that root.
A: 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 $x=1$.
A: Let $a=\sqrt[3]{2+\sqrt{5}}$ $b=\sqrt[3]{2-\sqrt{5}}$. Obviously $a b =-1$. 
Now let's forget what values of $a$ and $b$ and solve the system: 
$$
ab=-1\\
a+b=1
$$
They imply
$$
a^2-a-1=0
$$
namely $a=\frac {1\pm \sqrt{5}}{2}$. So $a=\frac {1+\sqrt{5}}{2}, b=\frac {1-\sqrt{5}}{2}$. Now taking $a^3,b^3$ we can easily get $2\pm\sqrt{5}$.
