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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?

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?

Grade:8

4 Answers

sahil
142 Points
6 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
d chandrasekhar
15 Points
5 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( k+1)^3  /abc
\small (a+b)(b+c)(c+a)= ( k+1)^3
d chandrasekhar
15 Points
5 years ago
a+b-c/c =a-b+c/b
b(a+b-c)=c(a-b+c)
ab+b^2-bc=ac-bc+c^2
ab-ac+b^2-c^2=0
a(b-c)+(b+c)(b-c)=0
(b-c)(a+b+c)=0
=> b-c=0
=> b=c
Similarly c=a
therefore a=b=c
(a+b)(b+c)(c+a)/abc=(a+a)(b+b)(c+c)/abc
2a.2b.2c/abc =8abc/abc
                   =8
ankit singh
askIITians Faculty 614 Points
3 years ago
 
a+b-c/c =a-b+c/b
b(a+b-c)=c(a-b+c)
ab+b^2-bc=ac-bc+c^2
ab-ac+b^2-c^2=0
a(b-c)+(b+c)(b-c)=0
(b-c)(a+b+c)=0
=> b-c=0
=> b=c
Similarly c=a
therefore a=b=c
(a+b)(b+c)(c+a)/abc=(a+a)(b+b)(c+c)/abc
2a.2b.2c/abc =8abc/abc
                   =8

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