Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ [duplicate]

Solution 1:

Let $s=a+b$ be our sum, where $a=\sqrt[3]{2+\sqrt{5}}$ and $b=\sqrt[3]{2-\sqrt{5}}$. Note that $$s^3=a^3+b^3+3ab(a+b)=a^3+b^3+3abs.$$ Thus since $a^3+b^3=4$ and $ab=\sqrt[3]{-1}=-1$, we have $s^3=4-3s$. This has the obvious root $s=1$ and no other real root.

Solution 2:

Would it help you to know that

${2\pm\sqrt5}=\left(\frac{1\pm\sqrt5}2\right)^3$ ?

Solution 3:

$(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} )^3 \\ =(\sqrt[3]{2 + \sqrt{5}})^3+(\sqrt[3]{2 - \sqrt{5}} )^3+3(\sqrt[3]{2 + \sqrt{5}} ) (\sqrt[3]{2 - \sqrt{5}} )(\sqrt[3]{2 + \sqrt{5}} +\sqrt[3]{2 - \sqrt{5}} ) $

S0 $s^3=4-3s$ From this we get S = 1