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.
Expanding both sides of the equation, we get
a^2 + 1 + b^2 + 1 + c^2 + 1 = (a + b + c) + (1/a + 1/b + 1/c)
Simplifying the right side, we get
a^2 + 1 + b^2 + 1 + c^2 + 1 = a + b + c + (1 + 1/a + 1/b + 1/c)
Subtracting 1 from both sides, we get
a^2 + b^2 + c^2 = a + b + c
Multiplying both sides by 3, we get
3a^2 + 3b^2 + 3c^2 = 3a + 3b + 3c
Using the equation abc = 13, we can write
3(a^2 + b^2 + c^2) = 3(a + b + c)/abc = 3(a + b + c)/13
Solving for a + b + c, we get
a + b + c = 3(a^2 + b^2 + c^2) * 13 = 39.
Therefore, the answer is 39.
(a + 1/b) (b + 1/c) (c + 1/a) = (1 + 1/a) (1 + 1/b) (1 + 1/c).........(1)
a x b x c =13...........................................................................(2)
a =-9, b= - 4 1/3, c=1/3
(-9 + -3/13) * (-13/3 + 1/(1/3)) * (1/(1/3) + 1/-9) =2.7350427
(1 + 1/-9) * (1 + 1/(-4 1/3)) * (1 + 1/(1/3)) ==2.7350427
a * b * c = -9 * - 4 1/3 * 1/3 = 13
a + b + c = - 9 + - 4 1/3 + 1/3 =- 13
