<s>[INST] 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.
 [/INST] \begin{align*}
&\left(10+\sqrt{108}\right)^{1/3}-\left(-10+\sqrt{108}\right)^{1/3}\\
&=\left(10+6\sqrt{3}\right)^{1/3}-\left(-10+6\sqrt{3}\right)^{1/3}\\
&=\left((1+\sqrt{3})^3\right)^{1/3}-\left((\sqrt{3}-1)^3\right)^{1/3}\\
&=(1+\sqrt{3})-(\sqrt{3}-1)\\
&=2
\end{align*}
The formula:
$$(A+B\sqrt{3})^3=A^3+3\sqrt{3}A^2B+9B^2A+3\sqrt{3}B^3$$ is useful.
</s>