Writing $\sqrt[\large3]2+\sqrt[\large3]4$ with nested roots Let $C\subset\Bbb R$ be the smallest set containing $0$ and closed under whole number addition/subtraction, whole number exponents, and whole number roots. That is, for all $c\in C$ and $n\in\Bbb N$, we have $c\pm n\in C$, $c^n\in C$, and $c^{1/n}\in C$.
We know that $\sqrt2+\sqrt3\in C$, since it's equal to $\sqrt{\sqrt{24}+5}$. Similarly, Ramanujan proved that:
$$\sqrt[\large3]{\sqrt[\large3]{1458}-9}-1=\sqrt[\large3]4-\sqrt[\large3]2$$
so $\sqrt[\large3]4-\sqrt[\large3]2\in C$. (Well, Ramanujan proved something equivalent.)
I found that $\left(\sqrt[\large3]2-1\right)^{-1}-1=\sqrt[\large3]2+\sqrt[\large3]4$, but I'm not allowed to use negative exponents/roots, so this doesn't prove that $\sqrt[\large3]2+\sqrt[\large3]4\in C$. So:

Is it true that $\sqrt[\large3]2+\sqrt[\large3]4\in C$? If not, why not?


Note
$$(\sqrt[3]2+\sqrt[3]4 -1)^2= 5- \sqrt[3]{4}
$$
which yields
$$\sqrt[3]2+\sqrt[3]4 =\sqrt{5-\sqrt[3]{4}}+1$$
