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Question Number 143899 by liberty last updated on 19/Jun/21

Answered by ajfour last updated on 19/Jun/21

let r^(10) =z t=z^4 −z^3 +z^2 −z+1 r^5 while ((r^5 −1)/(r−1))=0 let r=cos 72^° +isin 72° ⇒ but anyway r^5 =1 z=r^(10) =1 so t=r^(40) −r^(30) +r^(20) −r^(10) +1 =z^4 −z^3 +z^2 −z+1 =1−1+1−1+1 =1

letr10=zt=z4−z3+z2−z+1r5whiler5−1r−1=0letr=cos72°+isin72°⇒butanywayr5=1z=r10=1sot=r40−r30+r20−r10+1=z4−z3+z2−z+1=1−1+1−1+1=1

Answered by MJS_new last updated on 19/Jun/21

(x^4 −x^3 +x^2 −x+1)(x+1)=x^5 +1=0 ⇔ x^5 =−1 ⇒ x^(10) =x^(20) =x^(30) =x^(40) =1 ⇒ answer is 1

(x4−x3+x2−x+1)(x+1)=x5+1=0⇔x5=−1⇒x10=x20=x30=x40=1⇒answeris1

Answered by mr W last updated on 19/Jun/21

x^4 −x^3 +x^2 −x+1=0 (x+1)(x^4 −x^3 +x^2 −x+1)=0 x^5 +1=0 x^5 =−1 ⇒r^5 =−1 ∧r≠−1 r^(40) −r^(30) +r^(20) −r^(10) +1=((1+r^(50) )/(1+r^(10) ))=((1+(r^5 )^(10) )/(1+(r^5 )^2 )) =((1+(−1)^(10) )/(1+(−1)^2 ))=((1+1)/(1+1))=1

x4−x3+x2−x+1=0(x+1)(x4−x3+x2−x+1)=0x5+1=0x5=−1⇒r5=−1∧r≠−1r40−r30+r20−r10+1=1+r501+r10=1+(r5)101+(r5)2=1+(−1)101+(−1)2=1+11+1=1

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