What will you do?

Let $\color{#E81990}{A}= \sqrt[3]{2+\sqrt{5}}$ , $\color{#EC7300}{B}=\sqrt[3]{2-\sqrt{5}}$ .
And, $\Rightarrow\color{#D61F06}{S}=\color{#3D99F6}{A+B}=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Now, Cubing both sides.
$\Rightarrow \color{#D61F06}{S^3}=\color{#3D99F6}{(A+B)^3}$
$=\color{#3D99F6}{A^3+B^3}+3\color{#20A900}{AB}\color{#3D99F6}{(A+B)}$
$=\color{#3D99F6}{A^3+B^3}+\color{#20A900}{3AB}\color{#D61F06}{S}$ ...... $(\color{#20A900}{♧})$ [ Since , $\color{#3D99F6}{(A+B)}=\color{#D61F06}{S}$ ]
Thus,
$\Rightarrow \color{#3D99F6}{A^3+B^3}=(\sqrt[3]{2+\sqrt{5}})^{3}+(\sqrt[3]{2-\sqrt{5}})^{3}$
$=2+\sqrt{5}+2-\sqrt{5}$
$=\color{#302B94}{4}$

$\Rightarrow \color{#20A900}{3AB}=3(\sqrt[3]{2+\sqrt{5}})(\sqrt[3]{2-\sqrt{5}})$
$=3(\sqrt[3]{4-5})$
$=3(\sqrt[3]{-1})$
$=3×(-1)$
$=\color{#BA33D6}{-3}$
Now, putting $\color{#3D99F6}{(A^3+B^3)}=\color{#302B94}{4}$ and $\color{#20A900}{(3AB)}=\color{#BA33D6}{-3}$ in $(\color{#20A900}{♧})$ .
$\Rightarrow \color{#D61F06}{S^3}=\color{#302B94}{4}\color{#BA33D6}{-3}\color{#D61F06}{S}$
$\color{#D61F06}{S^3}\color{#BA33D6}{+3}\color{#D61F06}{S}\color{#302B94}{-4}=0$
$(\color{#D61F06}{S}-1)(\color{#D61F06}{S^2}+\color{#D61F06}{S}\color{#302B94}{+4})=0$
$\color{#D61F06}{S}-1=0$
$\color{#D61F06}{S}=\color{#624F41}{\boxed{1}}$

$\Rightarrow \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}=\color{#624F41}{\boxed{1}}.$

Note : There is a possibilty that $(\color{#D61F06}{S^2}+\color{#D61F06}{S}\color{#302B94}{+4}=0)$ ,but this is not possible because its roots are non-real complex numbers.And the surds given is real.

Let $y = \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}$ . Cubing it on both side, we get

$y^3 = (\sqrt[3]{2+\sqrt{5}})^3 + (\sqrt[3]{2-\sqrt{5}})^3 + 3( \sqrt[3]{2+\sqrt{5}})( \sqrt[3]{2-\sqrt{5}})(\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}) \\ y^3 = (2+\sqrt{5}) + (2-\sqrt{5}) + 3(\sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})})(y) \\ y^3 = 4 - 3y \\ y^3 +3y-4 =0$

By trial and error method, we get $y=1$ is a solution of this cubic equation. So, by synthetic division, we get

$y^3 + 3y -4 = (y-1)(y^2 +y +4)$

So, $y^3 + 3y -4 = 0 \implies y=1$ or $y^2 + y+ 4 = 0$ . But the roots of $y^2 + y +4 =0$ are non-real complex numbers, because the discriminant of the quadratic is negative. The surd given to us is real. So, the only possibility is $y=1$ .

Therefore, $\large \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} = \boxed{1}$

$\begin{aligned} \text{We note that: } (1\pm\sqrt{5})^3 & = 1\pm3(1^2)\sqrt{5}+3(1)(5) \pm 5\sqrt{5} \\ & = 16 \pm 8\sqrt{5} \\ \Rightarrow \left(\frac{1 \pm \sqrt{5}}{2}\right)^3 & = 2 \pm \sqrt{5} \end{aligned}$

$\begin{aligned} \text{Therefore, } \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} & = \sqrt[3]{\left(\frac{1 + \sqrt{5}}{2}\right)^3} + \sqrt[3]{\left(\frac{1 - \sqrt{5}}{2}\right)^3} \\ & = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} \\ & = \boxed{1} \end{aligned}$