Find the roots of $x^3 - 6x = 4$ This exercise is from the book Complex Analysis by Joseph Bak, and it says:
"Find the three roots of $x^{3}-6x=4$ by finding the three real-valued possibilities for $\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$".
I know that these numbers were found by Cardan's method, but I don't understand why they give these numbers, because I found three real roots by common methods.
Pd: the three real roots are $-2$, $1-\sqrt{3}$ and $1+\sqrt{3}$.

Use $2+2i=(-1+i)^3$ and $2-2i=(-1-i)^3.$
Now, let $\omega=-\frac{1}{2}+\frac{\sqrt3}{2}i.$
Thus, $\omega^2+\omega+1=0$ and $$(-1+i)+(-1-i),$$ $$(-1+i)w+(-1-i)w^2$$ and
$$(-1+i)w^2+(-1-i)w$$ give our real roots.
The first line I got by the following way:
$$2+2i=-2i(-1+i)=(-1+i)^2(-1+i)=(-1+i)^3$$ and
$$2-2i=2i(-1-i)=(-1-i)^2(-1-i)=(-1-i)^3.$$
