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?

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