What real numbers do algebraic numbers cover? Hardy and Wright mention ( though don't give a proof ) that any finite combination of real quadratic surds is an algebraic number. For example
$\sqrt{11+2\sqrt{7}}$. Are all finite combinations of cube root, fourth root ... $n^{th}$ root also algebraic ? such as $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$.

The usual ways to prove this is to use some basic linear algebra.
Assume $\alpha$ and $\beta$ are algebraic numbers. 
The field $\mathbb Q[\alpha][\beta]$ is a finite-dimensional vector space over $\mathbb Q$. Let $d$ be the dimension of the vector space. 
Then $1,\alpha+\beta,(\alpha+\beta)^2,\cdots,(\alpha+\beta)^d$ are $d+1$ elements of that vector space, and thus must be linearly dependent. But that means that $\alpha+\beta$ is the root of a rational polynomial of degree at most $d$.
The same for $\alpha\beta$.
More specifically, if $\alpha$ is the root of a rational polynomial of degree $d_1$ and $\beta$ is the root of a rational polynomial of degree $d_2$ then $\alpha+\beta$ and $\alpha\beta$ are both roots of a polynomial of degree at most $d_1d_2$.
Finally, if $\alpha$ is the root of the rational polynomial $p(x)$ then $\sqrt[n]\alpha$ is a root of $p(x^n)$.
So this means that $5+3\sqrt6$ is the root of a polynomial of degree $2$, and $3\sqrt[7]{5+3\sqrt6}$ is a root of a polynomial of degree $14$ and $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt6}}$ is the root of a polynomial of degree $42$ and finally $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$ is a root of a polynomial of degree $42\cdot 9=378$. There might be a polynomial of smaller degree, but we get this upper bound.
