<s>[INST] 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?
 [/INST] 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 $2\pm\sqrt{5}$.
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