Question-210786

Question Number 210786 by zhou0429 last updated on 19/Aug/24

Answered by Berbere last updated on 19/Aug/24

x^{9} + 1 = \underset{k = 0}{\prod^{8}} (x - e^{i π (\frac{1 + 2 k}{9})}) = p (x)

= Σ \int \frac{1}{(x - e^{i π (\frac{1 + 2 k}{9})}) p^{'} (e^{i π (\frac{1 + 2 k}{9})})} d x

= \underset{k = 0}{\sum^{\infty}} \frac{l n (x - e^{i π (\frac{1 + 2 k}{9})})}{P^{'} (e^{i π (\frac{1 + 2 k}{9})})} + c

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