Odpowiedź :
[tex]z \in \mathbb{C} \setminus \{0\}\\z^6 = -8\\z = \sqrt[6]{-8}[/tex]
[tex]z = \left| z \right| \left( \cos \varphi + i \sin \varphi \right) = 8 \left( \cos \pi + i \sin \pi \right)\\[/tex]
[tex]w_k = \sqrt[n]{ \left| z \right|} \left( \cos \frac{\varphi + 2 k \pi}{n} + i \sin \frac{\varphi + 2 k \pi}{n} \right)\\k \in \{0,\ 1,\ 2,\ 3,\ 4,\ 6 \}[/tex]
[tex]\sqrt[6]{ \left| -8 \right| } = \sqrt[6]{8} = 8^{\frac{1}{6} } = \left( 2^3 \right)^{\frac{1}{6} } = 2^{\frac{3\cdot1}{6} } = 2^{\frac{1}{2} } = \sqrt{2}[/tex]
[tex]z \in \{w_0,\ w_1,\ w_2,\ w_3,\ w_4,\ w_5 \}[/tex]
[tex]w_0 = \sqrt{2} \left( \cos \frac{\pi + 2 \cdot 0 \cdot \pi}{6} + i \sin \frac{\pi + 2 \cdot 0 \cdot \pi}{6} \right) = \sqrt{2} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) = \sqrt{2} \left( \frac{\sqrt{3} }{2} + \frac{1}{2} i \right) = \frac{\sqrt{6} }{2} + \frac{\sqrt{2} }{2}i[/tex]
[tex]w_1 = \sqrt{2} \left( \cos \frac{\pi + 2 \cdot 1 \cdot \pi}{6} + i \sin \frac{\pi + 2 \cdot 1 \cdot \pi}{6} \right) = \sqrt{2} \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) = \sqrt{2} \left( 0 + i \right) = \sqrt{2}i[/tex]
[tex]w_2 = \sqrt{2} \left( \cos \frac{\pi + 2 \cdot 2 \cdot \pi}{6} + i \sin \frac{\pi + 2 \cdot 2 \cdot \pi}{6} \right) = \sqrt{2} \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right) = \sqrt{2} \left( -\frac{\sqrt{3} }{2} \right + \frac{1}{2}i ) = -\frac{\sqrt{6} }{2} + \frac{\sqrt{2} }{2} i[/tex]
[tex]w_3 = \sqrt{2} \left( \cos \frac{\pi + 2 \cdot 3 \cdot \pi}{6} + i \sin \frac{\pi + 2 \cdot 3 \cdot \pi}{6} \right) = \sqrt{2} \left( \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right) = \sqrt{2} \left( - \frac{\sqrt{3} }{2} - \frac{1}{2}i \right) = - \frac{\sqrt{6} }{2} - \frac{\sqrt{2} }{2}i[/tex]
[tex]w_4 = \sqrt{2} \left( \cos \frac{\pi + 2 \cdot 4 \cdot \pi}{6} + i \sin \frac{\pi + 2 \cdot 4 \cdot \pi}{6} \right) = \sqrt{2} \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right) = \sqrt{2} \left( 0 - i) = -\sqrt{2} i[/tex]
[tex]w_5 = \sqrt{2} \left( \cos \frac{\pi + 2 \cdot 5 \cdot \pi}{6} + i \sin \frac{\pi + 2 \cdot 5 \cdot \pi}{6} \right) = \sqrt{2} \left( \cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6} \right) = \sqrt{2} \left( \frac{\sqrt{3} }{2} - \frac{1}{2} i \right) = \frac{\sqrt{6} }{2} - \frac{\sqrt{2} }{2}i[/tex]