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Question Number 13005 by 1630321995 last updated on 10/May/17
using De Moivre theorem solve the equation (x+1)^5 +(x−1)^5 =0
$${using}\:{De}\:{Moivre}\:{theorem}\:{solve}\:{the}\:{equation}\:\left({x}+\mathrm{1}\right)^{\mathrm{5}} +\left({x}−\mathrm{1}\right)^{\mathrm{5}} =\mathrm{0} \\ $$
Answered by mrW1 last updated on 10/May/17
(x+1)^5 −(1−x)^5 =0  (x+1)^5 =(1−x)^5   x+1=1−x  2x=0  ⇒x=0
$$\left({x}+\mathrm{1}\right)^{\mathrm{5}} −\left(\mathrm{1}−{x}\right)^{\mathrm{5}} =\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{5}} =\left(\mathrm{1}−{x}\right)^{\mathrm{5}} \\ $$$${x}+\mathrm{1}=\mathrm{1}−{x} \\ $$$$\mathrm{2}{x}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0} \\ $$

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