If \(abc= 13\) and \(\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)=\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\), find \(a +b+c\).
+21874 Multiplying out and simplifying the left hand side: (a + 1/b)(b + 1/c)(c + 1/a)
gives you: abc + (a + b + c) + (1/a + 1/b + 1/c) + 1/(abc)
Multiplying out and simplifying the right hand side: (1 + 1/a)(1 + 1/b)(1 + 1/c)
gives you: 1 + (1/a + 1/b + 1/c) + ( 1/(ab) + 1/(ac) + 1/(bc) ) + 1/(abc)
Setting these two equal to each other and subtracting both (1/a + 1/b + 1/c) and 1/(abc) from both sides,
leaves you with: abc + (a + b + c) = 1 + 1/(ab) + 1/(ac) + 1/(bc)
Rewrite 1/(ab) + 1/(ac) + 1/(bc) as (a + b + c)/(abc)
the equation becomes: abc + (a + b + c) = 1 + (a + b + c)/(abc)
Since abc = 13: 13 + (a + b + c) = 1 + (a + b + c)/13
Subtract 1: 12 + (a + b + c) = (a + b + c)/13
Multiply by 13: 156 + 13(a + b + c) = (a + b + c)
Simplify: 156 + 12(a + b + c) = 0
12(a + b + c) = -156
a + b + c = -13
