Simplifying the result formula for depressed Cubic After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$
After plugging into the formula
$$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}$$
where $m=6$ and $n=20$, we get
$$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}.$$
However, we notice that, without using Cardano's formula, that $x=2$ is the solution for the equation $x^3+6x=20.$
My question is: how does the equation $$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}$$ get simplified to $x=2$?
P.S. I understand that it was Niccolo Fontana who first figured out how to solve depressed cubic, to give one the proper credit.

Let $a^3 = 10+\sqrt{108}=10+6\sqrt{3}, b^3 = -10 + \sqrt{108}=-10+6\sqrt{3}\implies a^3-b^3 = 20 = (a-b)(a^2+ab+b^2)=(a-b)((a-b)^2 + 3ab)$. Observe that $(ab)^3 = 8 \implies ab = 2\implies (a-b)((a-b)^2 + 6) = 20\implies x^3+6x-20 = 0, x = a-b$. Using synthetic division we have: $x^3+6x-20 = 0\implies (x-2)((x+1)^2 + 9) = 0 \implies x = 2$ as claimed.
