Z^6 + 64 = 0
there are suppose to be 6 solutions
need help on how to find the roots
Are you familiar with de Moivre's theorem? The problem is pretty easy then. Alternatively, you can factor the equation [tex]Z^6 + 64=0[/tex] into
[tex](Z^3 - 8i)(Z^3+8i)[/tex] from which you can factor more.
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!
Write [itex]z^6 = r^6(cos 6\theta + isin 6\theta) [/itex] and solve.
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).
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