Writing $\sqrt[\large3]2+\sqrt[\large3]4$ with nested roots
$\sqrt[\large3]2+\sqrt[\large3]4$ is indeed in $C$: $$\sqrt{(\sqrt[\large3]{2} + 1)^2+3}=\sqrt[\large3]2+\sqrt[\large3]4$$
First, let's think about the process the other way around, instead of starting at $0$ and working our way to the number, start at the number and get to a whole number. Our goal is to get an expression with only one surd. After some messing around you realize that just addition and exponentiation isn't cutting it like it did for the other examples, so we must take a root. For convenience let $x=\sqrt[\large3]2$, and observe that $ax^2 + bx + c$ can represent every result of the operations allowed. So if we can get to a quadratic with an integer single root, we found a way to get rid off one of the surds.
$$(x^2+x)^2=x^4+2x^3+x^2=2x+4+x^2=x^2+2x+1+3=(x+1)^2+3$$ $$x^2+x=\sqrt{(x+1)^2+3}$$
$\sqrt[\large3]{6+\sqrt{(\sqrt[\large3]{864}+3)^2+135}}$ was the original solution, based on the quadratic ($x^2 + x +\frac{1}{4}$) my friend suggested (He came up with the idea of thinking about it as a quadratic with a single root). By the way, we can use the same method to get another identity: $$\sqrt{(\sqrt[\large3]{2} + 1)^2-5}=\sqrt[\large3]4-\sqrt[\large3]2$$
My follow-up questions are:
- $\sqrt[\large3]2+\sqrt[\large3]4$ can be expressed in similar way but without whole number exponentiation?
- Is $\sqrt[\large3]2+\sqrt[\large3]3$ in $C$? Is $\sqrt2+\sqrt[\large3]3$ in $C$?
Note $$(\sqrt[3]2+\sqrt[3]4 -1)^2= 5- \sqrt[3]{4} $$ which yields $$\sqrt[3]2+\sqrt[3]4 =\sqrt{5-\sqrt[3]{4}}+1$$