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Question Number 118990 by Algoritm last updated on 21/Oct/20

Answered by mr W last updated on 22/Oct/20

x≠1 ((x^5 −1)/(x−1))=0 ⇒x^5 =1 x^(20) +x^(10) =(x^5 )^4 +(x^5 )^2 =1^4 +1^2 =2

x≠1x5−1x−1=0⇒x5=1x20+x10=(x5)4+(x5)2=14+12=2

Answered by Bird last updated on 22/Oct/20

⇒((1−x^5 )/(1−x))=0 ⇒x^5 =1 and x≠1 the roots of z^(5 ) =1 are z_k =e^((i2kπ)/5) k∈[1,4] z_k ^(20) +z_k ^(10) =e^(8ikπ) +e^(4ikπ) =1+1=2

⇒1−x51−x=0⇒x5=1andx≠1therootsofz5=1arezk=ei2kπ5k∈[1,4]zk20+zk10=e8ikπ+e4ikπ=1+1=2

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