cube root simplification Wolfram Alpha correctly simplifies $\left(48\sqrt{3}+80\right)^{1/3}$ to $2+2\sqrt{3}.$ It can easily be verified that indeed $\left(2+2\sqrt{3}\right)^3=48\sqrt{3}+80.$ However, I would like to know if there is method to determine if such a simplification exists and, if it exists, if there is a method to find it? Also, related: Is there a way to simplify $\left(48\sqrt{3}+80\right)^{1/3} - \left(48\sqrt{3}-80\right)^{1/3}$? (Wolfram says it equals $4$, but I cannot derive (prove) it.) 
 A: One way to show if a simplification may exist is to look at the norm of your object. The norm of $a+b\sqrt c$ for a square-free $c$ is $a^2-cb^2$. This is multiplicative, meaning
$$norm(u\cdot v)=norm(u)\cdot norm(v)$$
As a result, the norm of a cube is the cube of the norm, and the norm of a cube root is the cube root of the norm. In your case, the norm of $80+48\sqrt 3$ is $80^2-3\cdot 48^2=-512$. If the norm were not a perfect cube, there would be no cube root of $80+48\sqrt 3$ that had integral coordinates. As it is, $-512$ is the cube of $-8$, so you can look for $a,b$ where $a^2-3b^2=-8$. And that could lead you to $a=\pm 2,b=\pm 2$ which leads to the answer. This method does not do the full job of simplification, but it helps. It definitely shows in many cases when such simplification is not possible.
Of course, as @Bernard implies, it would be easier to find the cube root by first factoring out an $8$ from $48\sqrt{3}+80$.
We can simplify $\left(48\sqrt{3}+80\right)^{1/3} - \left(48\sqrt{3}-80\right)^{1/3}$ by using the concept of the conjugate. There is an isomorphism from $a+b\sqrt c$ to $a-b\sqrt c$, meaning that if $a+b\sqrt c$ satisfies a polynomial with integral or rational coefficients, its conjugate also satisfies that polynomial.
Therefore, since we know that $\left(80+48\sqrt{3}\right)^{1/3}=2+2\sqrt{3}$ we also immediately know that $\left(80-48\sqrt{3}\right)^{1/3}=2-2\sqrt{3}$. This directly leads us to
$$\left(48\sqrt{3}+80\right)^{1/3} - \left(48\sqrt{3}-80\right)^{1/3}$$
$$\left(80+48\sqrt{3}\right)^{1/3} + \left(80-48\sqrt{3}\right)^{1/3}$$
$$=(2+2\sqrt3)+(2-2\sqrt3)$$
$$=4$$
A: For the related question: you can check $\bigl(2(\sqrt 3-1)\bigr)^3=8(6\sqrt 3-10)=48\sqrt 3-80$, hence 
$$\left(48\sqrt{3}+80\right)^{1/3} - \left(48\sqrt{3}-80\right)^{1/3}=(2+2\sqrt3)-(2\sqrt3-2)=4.$$
