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The roots of the equation $(x-1)^5=32(x+1)^5$ are
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Verified Answer
The correct answer is:
$\frac{1+2 e^{\frac{2 k \pi i}{5}}}{1-2 e^{\frac{2 k \pi i}{5}}}, k=1,2,3,4,5$
We have, $(x-1)^5=32(x+1)^5$
$\Rightarrow\left(\frac{x-1}{x+1}\right)^5=32 \Rightarrow \frac{x-1}{x+1}=(32)^{1 / 5} \Rightarrow \frac{x-1}{x+1}=2 e^{\frac{2 k \pi i}{5}}$
Apply componendo and dividendo
$\frac{2}{2 x}=\frac{1-2 e^{\frac{2 k \pi i}{5}}}{1+2 e^{\frac{2 k \pi i}{5}}} \Rightarrow x=\frac{1+2 e^{\frac{2 k \pi i}{5}}}{1-2 e^{\frac{2 k \pi i}{5}}}$
$\Rightarrow\left(\frac{x-1}{x+1}\right)^5=32 \Rightarrow \frac{x-1}{x+1}=(32)^{1 / 5} \Rightarrow \frac{x-1}{x+1}=2 e^{\frac{2 k \pi i}{5}}$
Apply componendo and dividendo
$\frac{2}{2 x}=\frac{1-2 e^{\frac{2 k \pi i}{5}}}{1+2 e^{\frac{2 k \pi i}{5}}} \Rightarrow x=\frac{1+2 e^{\frac{2 k \pi i}{5}}}{1-2 e^{\frac{2 k \pi i}{5}}}$
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