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 $1+\sqrt{3}$.

Let $x=\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$. Then,
$x^3=2+2i+2-2i+3(\sqrt[3]{2+2i})^2(\sqrt[3]{2-2i})+3(\sqrt[3]{2+2i})(\sqrt[3]{2-2i})^2$
$=4+3(\sqrt[3]{2+2i}+\sqrt[3]{2-2i})(\sqrt[3]{2+2i})(\sqrt[3]{2-2i})$
$=4+3x\sqrt[3]{(2+2i)(2-2i)}$
$=4+3x\sqrt[3]{8}$
$=4+6x$
$\therefore x^3-6x=4$
Therefore, solving for $x=\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$ can help you find the solutions.
