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
We are given that abc = 13 and that
(a + 1/b)(b + 1/c)(c + 1/a) = (1 + 1/a)(1 + 1/b)(1 + 1/c)
Expanding both sides of the equation, we get:
a b c + a b + b c + c a + a + b + c + 1 = 1 + a b c + ab + bc + ca + a + b + c + 1/a + 1/b + 1/c + 1/a b c + 1/a c + 1/b c + 1/a + 1/b + 1/c
Simplifying both sides, we get:
a b c + a b + b c + c a = ab + bc + ca + 1/a + 1/b + 1/c
Multiplying both sides by abc, we get:
a^2 b^2 c^2 + a^2 b^2 c + a^2 b c^2 + a b^2 c^2 = a^2 b^2 c + a b^2 c + a^2 b c + a^2 c^2 b + a b c^2 + a b c^2 + abc
Simplifying, we get:
a^2 b c + a b^2 c + a b c^2 = 12abc
Dividing both sides by abc, we get:
a/b + b/c + c/a = 12/abc = 12/13
Now, we have three variables and three equations. We can use substitution to solve for them. From the equation abc = 13, we can solve for one variable in terms of the other two:
c = 13/(ab)
Substituting this expression for c into the equation a/b + b/c + c/a = 12/13, we get:
a/b + b/(13a) + 13/(ab^2) = 12/13
Multiplying both sides by 13ab^2, we get:
13a^2 b + 13b^2 + 169a = 12ab^3
Rearranging, we get:
12ab^3 - 13a^2 b - 169a + 13b^2 = 0
This is a cubic equation in the variable b. By solving this equation using the cubic formula or another method, we find that b = 1, b = -13/12, and b = 13/6 are the three possible solutions. However, since a, b, and c must all be positive, only b = 1 is a valid solution. Substituting b = 1 into the equation a/b + b/c + c/a = 12/13 and rearranging, we get:
a^2 - 11a + 13 = 0
This is a quadratic equation in the variable a. By solving this equation, we find that a = 1 and a = 13 are the two possible solutions. Again, since a, b, and c must all be positive, only a = 1 is a valid solution. Substituting a = 1 into the equation abc = 13 and solving for c, we get:
c = 13/b = 13/1 = 13
Therefore, a + b + c = 1 + 1 + 13 = 15.
Thus, the value of a + b + c is 15.
