$x^6 +1 = 0 \ x^6 = -1 = (\cos \pi + i \sin \pi ) \ \text{From De Movire's Theorem} \ x = (\cos \pi + i \sin \pi ) ^ {\Big(\dfrac{1}{6}\Big)} = \Big(\cos \dfrac{(\pi + 2\pi k)}{6} +i \sin \dfrac{(\pi+2\pi k)}{6} \Big) \ (k = 0,1,2,3,4,5), \text{since a 6th degree polynomial can only have 6 roots.} \ ~ \ when \quad k=0, \ x= \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} = \dfrac{\sqrt 3}{2} + i \dfrac{1}{2} \ ~ \ when \quad k=1, \ x= \cos \dfrac{3\pi}{6} + i \sin \dfrac{3\pi}{6} = 0+1i = i \ ~ \ when \quad k=2, \ x= \cos \dfrac{5\pi}{6} + i \sin \dfrac{5\pi}{6} = \dfrac{-\sqrt 3}{2} + i\dfrac{1}{2} \ ~ \ when \quad k=3, \ x= \cos \dfrac{7\pi}{6} + i \sin \frac{7\pi}{6} = \dfrac{-\sqrt 3}{2}-i\dfrac{1}{2} \ ~ \ when \quad k=4, \ x= \cos \dfrac{9\pi}{6} + i \sin \dfrac{9\pi}{6} =\cos \dfrac{3\pi}{2} + i \sin \dfrac{3\pi}{2}= 0-1i =-i \ ~ \ when \quad k=5, \ x= \cos \dfrac{11\pi}{6} + i \sin \dfrac{11\pi}{6} = \dfrac{\sqrt 3}{2}-i\dfrac{1}{2}$ Hence, the 6 roots of $x^6+1=0$ are  ${x = i, -i, \dfrac{\sqrt 3}{2}+\dfrac{i}{2},-\dfrac{\sqrt 3}{2}+\dfrac{i}{2},\dfrac{\sqrt 3}{2}-\dfrac{i}{2},-\dfrac{\sqrt 3}{2}-\dfrac{i}{2} \quad } $