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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=?
x2+x+1=0,x4+x2+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
Anotherway(x2+x+1)2=02x4+x2+1+2x3+2x+2x2=0x4+x2+1+2x(x2+x+1)=0x4+x2+1+2x(0)=0x4+x2+1=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
x2+x+1=0=>x2+1=−x=>(x2+1)2=(−x)2=>x4+2x2+1=x2=>x4+x2+1=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
x2+x+1=0,x4+x2+1=?∙x2+x+1=0⇒x2+1=−x∙x4+x2+1=x4+2x2+1−x2=(x2+1)2−x2=(−x)2−x2=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
x2+x+1=0,x4+x2+1=?(x−1)(x2+x+1)=0x3−1=0x=1,ω,ω2∵x=1isrootofx−1∴ω,ω2aretherootsofgivenequation.Case1:x=ωx4+x2+1=ω4+ω2+1=ω3.ω+ω2+1=ω2+ω+1=0Case2:x=ω2x4+x2+1=(ω2)4+(ω2)2+1=ω8+ω4+1=(ω3)2.ω2+ω3.ω+1=ω2+ω+1=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
x2+x+1=0,x4+x2+1=?x4+x2+1=x4+2x2+1−x2=(x2+1)2−x2=(x2+x+1)(x2−x+1)=0×(x2−x+1)=0

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