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Question Number 218485 by Mamadi last updated on 11/Apr/25
solve the equation 1) X^6 −1=0 2) X^4 +X^2 +1=0
$${solve}\:{the}\:{equation} \\ $$$$\left.\mathrm{1}\right)\:\:\:{X}^{\mathrm{6}} −\mathrm{1}=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\:{X}^{\mathrm{4}} +{X}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$
Answered by Rasheed.Sindhi last updated on 11/Apr/25
(1) x^6 −1=0 (2) x^4 +x^2 +1=0 unnecessary (1)⇒(x+1)(x−1)(x^2 +x+1)(x^2 −x+1)=0 x+1=0∨ x−1=0 ∨ x^2 +x+1=0 ∨ x^2 +x+1=0 x=−1∨ x=1 ∨ x= ((−1±(√(1−4)))/2) ∨ x=((1±(√(1−4)))/2) x=−1∨ x=1 ∨ x= ((−1±i(√3))/2) ∨ x=((1±i(√3))/2) x=1,ω,ω^2 ,−1,−ω,−ω^2
$$\left(\mathrm{1}\right)\:\:\:{x}^{\mathrm{6}} −\mathrm{1}=\mathrm{0}\:\:\left(\mathrm{2}\right)\:\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\:{unnecessary} \\ $$$$\left(\mathrm{1}\right)\Rightarrow\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}+\mathrm{1}=\mathrm{0}\vee\:{x}−\mathrm{1}=\mathrm{0}\:\vee\:{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\:\vee\:{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=−\mathrm{1}\vee\:{x}=\mathrm{1}\:\vee\:{x}=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}}\:\vee\:{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}} \\ $$$${x}=−\mathrm{1}\vee\:{x}=\mathrm{1}\:\vee\:{x}=\:\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\vee\:{x}=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${x}=\mathrm{1},\omega,\omega^{\mathrm{2}} ,−\mathrm{1},−\omega,−\omega^{\mathrm{2}} \\ $$

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