Taking the cube root of a sum of radicals I am wondering how to derive the following simplification without knowing it beforehand:
$$^3\sqrt{10 + 6\sqrt{3}} = 1 + \sqrt{3}$$
After the fact, it is easy to verify algebraically. The problem arose when applying Cardano's method to solve
$$y^3 + 6y = 20$$
I was able to derive a similar but less complicated simplification,
$$\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}$$
by assuming that $3 + 2\sqrt{2} = (a + b\sqrt{2})^2$, expanding, equating coefficients of $\sqrt{2}$, and solving for $a$ and $b$ using the quadratic formula. However, using the same method, i.e. assuming that $10 + 6\sqrt{3} = (a + b\sqrt{3})^3$, expanding, and equating coefficients of $\sqrt{3}$ yields cubic equations in $a$ and $b$:
$$a^3 + 9ab^2 = 10, a^2b + b^3 = 2$$
Guess-and-check yields $a = 1$ and $b = 1$, but I would prefer a more systematic method of solution. I did not use the cubic formula again, figuring that this would probably yield another nested radical.
According to Wolfram|Alpha, $10 + 6\sqrt{3} = 1 + 3\sqrt{3} + 9 + 3\sqrt{3} = 1 + 3\sqrt{3} + 3(\sqrt{3})^2 + (\sqrt{3})^3 = (1 + \sqrt{3})^3$. However, I'm not sure how I would arrive at that chain of reasoning except by chance or by using Wolfram|Alpha.

$N(10+6\sqrt{3})=10^2-3\cdot6^2=-8$. So in $\mathbb Z[\sqrt3]$, it is possible that $10+6\sqrt{3}$ is a perfect cube. The only possible solution to $\sqrt[3]{10+6\sqrt{3}} $ will be one with norm $-2$. That yields possible cube roots $(1\pm \sqrt{3})(2\pm \sqrt{3})^k$ for some $k$. 
