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1-a-1-b-1-c-1-a-b-c-Prove-that-1-a-5-1-b-5-1-c-5-1-a-5-b-5-c-5-1-a-b-c-5-

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Question Number 192437 by MATHEMATICSAM last updated on 18/May/23
(1/a) + (1/b) + (1/c) = (1/(a + b + c)) . Prove that (1/a^5 ) + (1/b^5 ) + (1/c^5 ) = (1/(a^5 + b^5 + c^5 )) = (1/((a + b + c)^5 ))
1a+1b+1c=1a+b+c.Provethat1a5+1b5+1c5=1a5+b5+c5=1(a+b+c)5
Answered by Frix last updated on 18/May/23
(1/a)+(1/b)+(1/c)=(1/(a+b+c)) Transforming (a+b)(a+c)(b+c)=0 ⇒ Only true if a=−b∨a=−c∨b=−c Because of symmetry let b=−c ⇒ (1/a)+(1/b)+(1/c)=(1/(a+b+c))=(1/a) ⇒ (1/a^(2n+1) )+(1/b^(2n+1) )+(1/c^(2n+1) )=(1/(a^(2n+1) +b^(2n+1) +c^(2n+1) ))= =(1/((a+b+c)^(2n+1) ))=(1/a^(2n+1) ) ∀n∈N
1a+1b+1c=1a+b+cTransforming(a+b)(a+c)(b+c)=0⇒Onlytrueifa=−b∨a=−c∨b=−cBecauseofsymmetryletb=−c⇒1a+1b+1c=1a+b+c=1a⇒1a2n+1+1b2n+1+1c2n+1=1a2n+1+b2n+1+c2n+1==1(a+b+c)2n+1=1a2n+1∀n∈N

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