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Question Number 174709 by AgniMath last updated on 09/Aug/22

Commented by MJS_new last updated on 09/Aug/22

±1?

±1?

Commented by infinityaction last updated on 09/Aug/22

if a ≠b≠c b +(1/c) = a+(1/b) a−b = ((b−c)/(bc)) ⇒ bc = ((b−c)/(a−b)) similarly ac = ((c−a)/(b−c)) ab = ((a−b)/(c−a)) (abc)^2 = ((b−c)/(a−b))×((a−c)/(b−c))×((a−c)/(c−a)) abc = ±1

ifa≠b≠cb+1c=a+1ba−b=b−cbc⇒bc=b−ca−bsimilarlyac=c−ab−cab=a−bc−a(abc)2=b−ca−b×a−cb−c×a−cc−aabc=±1

Commented by AgniMath last updated on 09/Aug/22

For answer go to: https://math.agnibho.co.in/2022/08/q12.html

Answered by behi834171 last updated on 09/Aug/22

a+(1/b)=b+(1/c)=c+(1/a)=k a+b+c≠0 ⇒a+b=bk,b+c=ck,c+a=ak ⇒2(a+b+c)=k(a+b+c)⇒k=2 a+b=2b⇒a=b=c ⇒a+(1/a)=2⇒a^2 −2a+1=0 ⇒a=b=c=1 .■

a+1b=b+1c=c+1a=ka+b+c≠0⇒a+b=bk,b+c=ck,c+a=ak⇒2(a+b+c)=k(a+b+c)⇒k=2a+b=2b⇒a=b=c⇒a+1a=2⇒a2−2a+1=0⇒a=b=c=1.◼

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