precal hw on complex numbers

[MercuryRising]

Z^6 + 64 = 0
there are suppose to be 6 solutions :confused:
need help on how to find the roots

[vsage]

Are you familiar with de Moivre's theorem? The problem is pretty easy then. Alternatively, you can factor the equation Z^6 + 64=0 into
(Z^3 - 8i)(Z^3+8i) from which you can factor more.

[MercuryRising]

opps, sorry, i must 've made myself unclear, the goal is to find the 6 values of z, but thanks for heling on the roots! :smile:

[Gokul43201]

Write z^6 = r^6(cos 6\theta + isin 6\theta) and solve.

[nolachrymose]

opps, sorry, i must 've made myself unclear, the goal is to find the 6 values of z, but thanks for heling on the roots! :smile:

Vsage's point was that, from there, you can factor further, thus getting the six linear roots (hint: think about sum and difference of cubes).