Find $$$\sqrt[3]{-8}$$$

Find $$$\sqrt[3]{-8}$$$.

The polar form of $$$-8$$$ is $$$8 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right)$$$ (for steps, see polar form calculator).

According to the De Moivre's Formula, all $$$n$$$-th roots of a complex number $$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$ are given by $$$r^{\frac{1}{n}} \left(\cos{\left(\frac{\theta + 2 \pi k}{n} \right)} + i \sin{\left(\frac{\theta + 2 \pi k}{n} \right)}\right)$$$, $$$k=\overline{0..n-1}$$$.

We have that $$$r = 8$$$, $$$\theta = \pi$$$, and $$$n = 3$$$.

• $$$k = 0$$$: $$$\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 0}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 0}{3} \right)}\right) = 2 \left(\cos{\left(\frac{\pi}{3} \right)} + i \sin{\left(\frac{\pi}{3} \right)}\right) = 1 + \sqrt{3} i$$$
• $$$k = 1$$$: $$$\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 1}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 1}{3} \right)}\right) = 2 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right) = -2$$$
• $$$k = 2$$$: $$$\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 2}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 2}{3} \right)}\right) = 2 \left(\cos{\left(\frac{5 \pi}{3} \right)} + i \sin{\left(\frac{5 \pi}{3} \right)}\right) = 1 - \sqrt{3} i$$$

$$$\sqrt[3]{-8} = 1 + \sqrt{3} i\approx 1 + 1.732050807568877 i$$$A

$$$\sqrt[3]{-8} = -2$$$A

$$$\sqrt[3]{-8} = 1 - \sqrt{3} i\approx 1 - 1.732050807568877 i$$$A