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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=Π_(k=0) ^8 (x−e^(iπ(((1+2k)/9))) )=p(x)  =Σ∫(1/((x−e^(iπ(((1+2k)/9))) )p′(e^(iπ(((1+2k)/9))) )))dx  =Σ_(k=0) ^∞ ((ln(x−e^(iπ(((1+2k)/9))) ))/(P′(e^(iπ(((1+2k)/9))) )))+c
$${x}^{\mathrm{9}} +\mathrm{1}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{8}} {\prod}}\left({x}−{e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right)={p}\left({x}\right) \\ $$$$=\Sigma\int\frac{\mathrm{1}}{\left({x}−{e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right){p}'\left({e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right)}{dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{ln}\left({x}−{e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right)}{{P}'\left({e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right)}+{c} \\ $$

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