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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 $a+b+c$.

 Jul 24, 2023
 #1
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abc = 13

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

 

(ab + 1)(bc + 1)(ca + 1) = (a + 1)(b + 1)(c + 1)

 

(abc)^2 + abc × (a + b + c) + (ab + bc + ca) + 1

 

= abc + (ab + bc + ca) + (a + b + c) + 1

 

(a + b + c) × (13 — 1) = 13 — 13^2

 

Answer, a + b + c

 

= 13 × (1 — 13) / (13 — 1)

 

a + b + c = –13

 Jul 24, 2023
 #4
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That is incorrect.

Guest Jul 26, 2023
 #2
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Simplify [(a + b^(-1)) (b + c^(-1)) (a^(-1) + c) == (1 + a^(-1)) (1 + b^(-1)) (1 + c^(-1))]


((-1 + a b c) (a + b + c + a b c))/(a b c) == 0


(( -1 +  13) (a + b + c + 13)) / 13 ==0, solve for a + b + c


12/13(a + b +c + 13)==0
Multiply both sides by 13/12


(a + b + c + 13)==0
Subtract 13 from both sides


a + b + c == - 13

 Jul 24, 2023
 #3
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Hi! It looks like you are cheating on your ao ps homework! Please post it on the message board and not here! Thanks!

 Jul 24, 2023

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