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

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

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.

• Thanks for answer Sep 11, 2020 at 15:50

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*}

• Thanks for your answer Sep 23, 2020 at 20:51