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`        If a b c are positive real numbers such that a+b-c/c=a-b+c/b=-a+b+c/a, find the value of (a+b) (b+c) (c+a) /abc?`
2 years ago

```							a+b-c/c=a-b+c/b=-a+b+c/a, find the value of (a+b) (b+c) (c+a) /abc?First take first and second eq a+b-c/c=a-b+c/b we get ba+b^2-bc=ca-bc+c^2 solve by cancelling and dividing by (b-c) you get b+c=-a similarly with taking eq second and third you get (a+b)=-c and from third and first (c+a)=-b now (a+b)(b+c)(c+a)/abc =-a×-b×-c/abc=-abc/abc=-1 Hence required answer is -1
```
2 years ago
```							a+b-c/c=a-b+c/b=-a+b+c/a = k(constant)then a+b-c=kc =>a+b=kc+c =>c(k+1)                           b+c = a(k+1)                           c+a = b(k+1)therefore (a+b)(b+c)(c+a)/abc= abc  /abc
```
one year ago
```							a+b-c/c =a-b+c/bb(a+b-c)=c(a-b+c)ab+b^2-bc=ac-bc+c^2ab-ac+b^2-c^2=0a(b-c)+(b+c)(b-c)=0(b-c)(a+b+c)=0=> b-c=0=> b=cSimilarly c=atherefore a=b=c(a+b)(b+c)(c+a)/abc=(a+a)(b+b)(c+c)/abc2a.2b.2c/abc =8abc/abc                   =8
```
one year ago
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