Express this sum of radicals as an integer? I have read somewhere that the radical $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$ and I don't understand it. How do you solve this(when the RHS is unknown)?

Let $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$. Write $a=2+\sqrt{5}$ and $b=2-\sqrt{5}$. Then
$$
x^3=
a+b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2}
=
a+b + 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})
=4-3x
$$
Now, the derivative of $x^3+3x-4$ is always positive, which means there is only one real root. By inspection, $x=1$ is that root.
