If a-b/c + b-c/a + c+a/b = 1 then show that 1/a = 1/b + 1/c
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(a-b)/c +(b-c)/a +(c+a)/b=1
(a-b)/c +(b-c)/a +(c+a)/b=1+1-1
(a-b)/c+1 +(b-c)/a-1 +(c+a)/b-1=0
{(a-b)/c+1} +{(b-c)/a-1} +{(c+a)/b-1}=0
(a-b+c)/c +(b-c-a)/a +(c+a-b)/b=0
(a-b+c)/c -(-b+c+a)/a +(c+a-b)/b=0
(a-b+c)/c -(a-b+c)/a +(a-b+c)/b=0
(a-b+c)(1/c-1/a+1/b)=0
so {1/c-1/a+1/b}=0
=>1/a=1/b+1/c.
(a-b)/c +(b-c)/a +(c+a)/b=1+1-1
(a-b)/c+1 +(b-c)/a-1 +(c+a)/b-1=0
{(a-b)/c+1} +{(b-c)/a-1} +{(c+a)/b-1}=0
(a-b+c)/c +(b-c-a)/a +(c+a-b)/b=0
(a-b+c)/c -(-b+c+a)/a +(c+a-b)/b=0
(a-b+c)/c -(a-b+c)/a +(a-b+c)/b=0
(a-b+c)(1/c-1/a+1/b)=0
so {1/c-1/a+1/b}=0
=>1/a=1/b+1/c.
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