How to reduce the radical? I was reading a pdf on Cardano's method of solving for the roots of a cubic polynomial, when I noticed an example.
$$x^3+6x-20=0$$
On solving, I got 
$$x=\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}$$
I went through a calculator and it gave the answer $2$.
My question is how can we prove 
$$\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}=2$$ 
I tried to simplify this radical, but arrive at the same cubic polynomial. No further advancement is taking place in my solving.
Another similar interesting cubic polynomial is
$$x^3-15x-4=0$$ It has the root
$$x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}$$
and read that 
$$\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}=4$$ I hope someone will help me.
 A: As  $\displaystyle10+\sqrt{108}=10+6\sqrt3,$
we can write $\displaystyle10+6\sqrt3=(a+b\sqrt3)^3$ where $a,b$ are real rationals
$\displaystyle\implies10+6\sqrt3=a^3+2a^2\cdot b\sqrt3+3a(b\sqrt3)^2+(b\sqrt3)^3$
$\displaystyle\implies10+6\sqrt3= a^3+9ab^2+3(a^2b+b^3)\sqrt3$
Comparing the rational & the irrational parts $\displaystyle 10=a^3+9ab^2\  \ \ \ (1)$ and $\displaystyle a^2b+b^3=2$
$\displaystyle\implies5(a^2b+b^3)=a^3+9ab^2\iff a^3-5a^2b+9ab^2-5b^3=0$
Clearly, $a=b$ is a solution
From $\displaystyle(1),10a^3=10\iff a^3=1\iff a=1$ as $a$ is real
So, from $(1)b^2=1$ and  from $(2),b^3+b-2=0$ the common root being $1$
A: Another way to find out the solutions is the following. For the fundamental theorem of algebra the solutions are three. You can easily see that $x^3+6x-20=(x-2)(x^2+2x+10)$ than there is one real solution $x_1=2$. The other two solutions are complex, in accordance with the Cardano's formulas
A: The following way does show that your value equals $2$.
Letting $f(x)=x^3+6x-20,$ since we have
$$f^\prime(x)=3x^2+6\gt0.$$
Hence, $f(x)=0$ has exactly one real solution. 
Hence, we know that the following holds :
$$\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}=2.$$
