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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 $1+\sqrt{3}$.

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3 Answers 3

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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 $$2-2i=2i(-1-i)=(-1-i)^2(-1-i)=(-1-i)^3.$$

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  • $\begingroup$ Thanks for answer, but I don't understand the first part, which is the origin of -1+i and - 1-i? $\endgroup$ Sep 11, 2020 at 5:39
  • $\begingroup$ @Brigitte Eliana I added something. See now. $\endgroup$ Sep 11, 2020 at 5:46
  • $\begingroup$ Thanks, I'm going to see $\endgroup$ Sep 11, 2020 at 15:51
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    $\begingroup$ @Brigitte Eliana You are welcome! $\endgroup$ Sep 11, 2020 at 16:24
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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.

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  • $\begingroup$ Thanks for answer $\endgroup$ Sep 11, 2020 at 15:50
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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:

$$\begin{align*}\sqrt[3]{2+2i}+\sqrt[3]{2-2i} &= \sqrt[3]{2\sqrt{2}\left(\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}\right)}+\sqrt[3]{2\sqrt{2}\left(\dfrac{1}{\sqrt{2}}-i\dfrac{1}{\sqrt{2}}\right)}\\ \\ &= \left[2\sqrt{2}e^{i\left(\frac{\pi}{4}+2\pi k\right)}\right]^{\frac{1}{3}}+\left[2\sqrt{2}e^{-i\left(\frac{\pi}{4}+2\pi k\right)}\right]^{\frac{1}{3}} \quad k \in \{0,1,2\}\\ \\ &= \sqrt{2}\left[e^{i\left(\frac{\pi}{12}+\frac{2\pi}{3}k\right)}+e^{-i\left(\frac{\pi}{12}+\frac{2\pi}{3}k\right)}\right]\quad k \in \{0,1,2\} \\ \\ &= 2\sqrt{2}\cos\left(\frac{\pi}{12}+\frac{2\pi}{3}k\right) \quad k \in \{0,1,2\} \\ \\ &= 2\sqrt{2}\left[\cos\left(\frac{\pi}{12}\right)\cos\left(\frac{2\pi}{3}k\right) - \sin\left(\frac{\pi}{12}\right)\sin\left(\frac{2\pi}{3}k\right)\right] \quad k \in \{0,1,2\}\\ \\ &= \left(1+\sqrt{3}\right)\cos\left(\frac{2\pi}{3}k\right) -\left(\sqrt{3}-1\right) \sin\left(\frac{2\pi}{3}k\right) \quad k \in \{0,1,2\}\\ \\ &= \left\{1+\sqrt{3},-2 ,1-\sqrt{3}\right\}\\ \end{align*}$$

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  • $\begingroup$ Thanks for your answer $\endgroup$ Sep 23, 2020 at 20:51

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