Question

Розв’язати рівняння Х^3=-2

Answer 1

$x^3=-2\Rightarrow x=-\sqrt[3]{2}$

$\sqrt[3]{-2}=\sqrt[3]{2(\cos \pi+i\sin \pi)}=\sqrt[3]{2}\left ( \left ( \cos \frac{\pi +2\pi k}{3} \right )+i\sin \left ( \frac{\pi+2\pi k}{3} \right ) \right ),k\in 0,1,2$$k=0\Rightarrow x=\sqrt[3]{2}\left ( \cos \frac{\pi}{3}+i\sin \frac{\pi}{3} \right )=\frac{\sqrt[3]{2}}{2}+\frac{\sqrt{3}\sqrt[3]{2}}{2}i$$k=1\Rightarrow x=\sqrt[3]{2}\left ( \cos \pi+i\sin \pi \right )=-\sqrt[3]{2}$$k=2\Rightarrow x=\sqrt[3]{2}\left ( \cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3} \right )=\frac{\sqrt[3]{2}}{2}-\frac{\sqrt{3}\sqrt[3]{2}}{2}i$