Using de Moivre's theorem to solve t³ - 3t² - 3t + 1 = 0

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Well... I think so...
 
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sooyong94 said:
Well... I think so...

Can't think why not. You've found three roots with the trig identity that you correctly got through deMoivre. They all work. Cubic equations have a maximum of three roots. We've disposed of the idea t=1 is one of them. Can't think of what else could go wrong.
 
sooyong94 said:
Thanks then. :D

Very welcome.