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Resolved.

 Jun 6, 2020
edited by Guest  Jun 6, 2020
edited by Possible  Jun 6, 2020
 #1
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(a + 1/b)(b + 1/c)(c + 1/a)

=  [ (a + 1/b)(b + 1/c) ] (c + 1/a)

=  [ ab + a/c + 1 + 1/bc ] (c + 1/a)

=  abc + b + a + 1/c + c + 1/a + 1/b + 1/abc

=  abc + (a + b + c) + (1/a + 1/b + 1/c) + 1/abc

 

(1 + 1/a)(1 + 1/b)(1 + 1/c)

=  [ (1 + 1/a)(1 + 1/b) ]·(1 + 1/c)

=  [ 1 + 1/b + 1/a + 1/ab ]·(1 + 1/c)

=  1 + 1/c + 1/b + 1/bc + 1/a + 1/ac + 1/ab + 1/abc

=  1 + (1/a + 1/b + 1/c) + (1/ab + 1/ac + 1/bc) + 1/abc

 

Setting these two equal to each other:

abc + (a + b + c) + (1/a + 1/b + 1/c) + 1/abc  =  1 + (1/a + 1/b + 1/c) + (1/ab + 1/ac + 1/bc) + 1/abc

Subtracting like terms:

abc + (a + b + c)  =  1 + (1/ab + 1/ac + 1/bc)

Rewriting:

abc + (a + b + c)  =  1 + (c/abc + b/abc + a/abc)

Factoring:

abc + (a + b + c)  =  1 + (a + b + c)/abc

Since abc = 13:

13 + (a + b + c)  =  1 + (a + b + c)/13

Subtracting:

12 + (a + b + c)  =  (a + b + c)/13

Multiplying both sides by 13:

156 + 13(a + b + c)  =  (a + b + c)

Subtracting (a + b + c) from both sides:

156 + 12(a + b + c)  =  0

Subtracting 156 from both sides:

12(a + b + c)  =  -156

Dividing:

a + b + c  =  -13

 Jun 6, 2020

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