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Question Number 218099 by Rasheed.Sindhi last updated on 29/Mar/25

x^2 +x+1=0 , x^4 +x^2 +1=?

$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\:,\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=? \\ $$

Answered by Rasheed.Sindhi last updated on 29/Mar/25

Another way (x^2 +x+1)^2 =0^2 x^4 +x^2 +1+2x^3 +2x+2x^2 =0 x^4 +x^2 +1+2x(x^2 +x+1)=0 x^4 +x^2 +1+2x(0)=0 x^4 +x^2 +1=0

$$\mathrm{Another}\:\mathrm{way} \\ $$$$\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{x}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{x}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$

Answered by profcedricjunior last updated on 29/Mar/25

x^2 +x+1=0=>x^2 +1=−x =>(x^2 +1)^2 =(−x)^2 =>x^4 +2x^2 +1=x^2 =>x^4 +x^2 +1=0

$$\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{x}}+\mathrm{1}=\mathrm{0}=>\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}=−\boldsymbol{{x}} \\ $$$$=>\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} =\left(−\boldsymbol{{x}}\right)^{\mathrm{2}} \\ $$$$=>\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}=\boldsymbol{{x}}^{\mathrm{2}} =>\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$

Answered by ArshadS last updated on 29/Mar/25

x^2 +x+1=0 , x^4 +x^2 +1=? •x^2 +x+1=0⇒x^2 +1=−x •x^4 +x^2 +1=x^4 +2x^2 +1−x^2 =(x^2 +1)^2 −x^2 =(−x)^2 −x^2 =0

$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\:,\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=? \\ $$$$\bullet{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} +\mathrm{1}=−{x} \\ $$$$\bullet{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}−{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(−{x}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{0} \\ $$

Answered by ArshadS last updated on 30/Mar/25

x^2 +x+1=0 , x^4 +x^2 +1=? (x−1)(x^2 +x+1)=0 x^3 −1=0 x=1,ω,ω^2 ∵x=1 is root of x−1 ∴ ω,ω^2 are the roots of given equation. Case1: x=ω x^4 +x^2 +1=ω^4 +ω^2 +1 =ω^3 .ω+ω^2 +1=ω^2 +ω+1=0 Case2: x=ω^2 x^4 +x^2 +1=(ω^2 )^4 +(ω^2 )^2 +1 =ω^8 +ω^4 +1=(ω^3 )^2 .ω^2 +ω^3 .ω+1 =ω^2 +ω+1=0

$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\:,\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=? \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{1},\omega,\omega^{\mathrm{2}} \\ $$$$\because{x}=\mathrm{1}\:{is}\:{root}\:{of}\:{x}−\mathrm{1} \\ $$$$\therefore\:\omega,\omega^{\mathrm{2}} {are}\:{the}\:{roots}\:{of}\:{given}\:{equation}. \\ $$$${Case}\mathrm{1}:\:{x}=\omega \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=\omega^{\mathrm{4}} +\omega^{\mathrm{2}} +\mathrm{1} \\ $$$$=\omega^{\mathrm{3}} .\omega+\omega^{\mathrm{2}} +\mathrm{1}=\omega^{\mathrm{2}} +\omega+\mathrm{1}=\mathrm{0} \\ $$$${Case}\mathrm{2}:\:{x}=\omega^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=\left(\omega^{\mathrm{2}} \right)^{\mathrm{4}} +\left(\omega^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{1} \\ $$$$=\omega^{\mathrm{8}} +\omega^{\mathrm{4}} +\mathrm{1}=\left(\omega^{\mathrm{3}} \right)^{\mathrm{2}} .\omega^{\mathrm{2}} +\omega^{\mathrm{3}} .\omega+\mathrm{1} \\ $$$$=\omega^{\mathrm{2}} +\omega+\mathrm{1}=\mathrm{0} \\ $$

Answered by MATHEMATICSAM last updated on 30/Mar/25

x^2 +x+1=0 , x^4 +x^2 +1=? x^4 + x^2 + 1 = x^4 + 2x^2 + 1 − x^2 = (x^2 + 1)^2 − x^2 = (x^2 + x + 1)(x^2 − x + 1) = 0 × (x^2 −x + 1) = 0

$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\:,\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=? \\ $$$${x}^{\mathrm{4}} \:+\:{x}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$=\:{x}^{\mathrm{4}} \:+\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{1}\:−\:{x}^{\mathrm{2}} \\ $$$$=\:\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)^{\mathrm{2}} \:−\:{x}^{\mathrm{2}} \\ $$$$=\:\left({x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{1}\right) \\ $$$$=\:\mathrm{0}\:×\:\left({x}^{\mathrm{2}} \:−{x}\:+\:\mathrm{1}\right) \\ $$$$=\:\mathrm{0} \\ $$

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