Strange surds!

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 $\displaystyle x=(2+\sqrt{5})^{\frac{1}{3}}+(2-\sqrt{5})^{\frac{1}{3}} \implies x-(2+\sqrt{5})^{\frac{1}{3}}-(2-\sqrt{5})^{\frac{1}{3}}=0$

So we have $\displaystyle x^3-(2+\sqrt{5})-(2-\sqrt{5}) =3x(2^2-5)^{\frac{1}{3}} \implies x^3+3x-4=0$ , Since $x=1$ is a root of the equation and the other roots are non-real as $x^2+x+4=0$ has no real solutions it follows that $1=(2+\sqrt{5})^{\frac{1}{3}}+(2-\sqrt{5})^{\frac{1}{3}}$

Since integer solution is expected, we can assume $2 \pm \sqrt 5 = (a \pm b\sqrt 5)^3$ , then we have $\sqrt[3]{2 + \sqrt 5} + \sqrt[3]{2 - \sqrt 5}$ $= \sqrt[3]{(a + b\sqrt 5)^3} + \sqrt[3]{(a - b\sqrt 5)^3}$ $= a + b\sqrt 5 + a - b\sqrt 5 = 2a$ . We have:

$\begin{aligned} (a + b\sqrt 5)^3 & = 2 + \sqrt 5 \\ a^3 + 3a^2b\sqrt 5 + 15ab^2 + 5b^3\sqrt 5 & = 2 + \sqrt 5 \end{aligned}$

Equating the rational and irrational parts we have: $\begin{cases} a^3 + 15ab^2 = 2 \\ 3a^2b + 5b^3 = 1 \end{cases} \implies a = b = \dfrac 12$ .

$\implies \sqrt[3]{2 + \sqrt 5} + \sqrt[3]{2 - \sqrt 5} = 2a = 2 \times \dfrac 12 = \boxed{1}$