<s>[INST] 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 $1+\sqrt{3}$.
 [/INST] I find exponential form useful for a case known to have 3 real roots. One can get to a real expression with $\cos\left(\frac{\theta}{3}+\frac{2\pi}{3}k\right)$ and keep everything real from that point forward:
$$\begin{align*}\sqrt[3]{2+2i}+\sqrt[3]{2-2i} &= \sqrt[3]{2\sqrt{2}\left(\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}\right)}+\sqrt[3]{2\sqrt{2}\left(\dfrac{1}{\sqrt{2}}-i\dfrac{1}{\sqrt{2}}\right)}\\
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&= \left[2\sqrt{2}e^{i\left(\frac{\pi}{4}+2\pi k\right)}\right]^{\frac{1}{3}}+\left[2\sqrt{2}e^{-i\left(\frac{\pi}{4}+2\pi k\right)}\right]^{\frac{1}{3}} \quad k \in \{0,1,2\}\\
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&= \sqrt{2}\left[e^{i\left(\frac{\pi}{12}+\frac{2\pi}{3}k\right)}+e^{-i\left(\frac{\pi}{12}+\frac{2\pi}{3}k\right)}\right]\quad k \in \{0,1,2\} \\
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&= 2\sqrt{2}\cos\left(\frac{\pi}{12}+\frac{2\pi}{3}k\right) \quad k \in \{0,1,2\} \\
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&= 2\sqrt{2}\left[\cos\left(\frac{\pi}{12}\right)\cos\left(\frac{2\pi}{3}k\right) - \sin\left(\frac{\pi}{12}\right)\sin\left(\frac{2\pi}{3}k\right)\right] \quad k \in \{0,1,2\}\\
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&= \left(1+\sqrt{3}\right)\cos\left(\frac{2\pi}{3}k\right) -\left(\sqrt{3}-1\right) \sin\left(\frac{2\pi}{3}k\right) \quad k \in \{0,1,2\}\\
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&= \left\{1+\sqrt{3},-2 ,1-\sqrt{3}\right\}\\
\end{align*}$$
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