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0
135
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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.

Oct 6, 2019

#1
+8853
+2

Let's start by expanding both sides and combining like terms:

$$\begin{array}{rcl} \left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)&=& \left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\\~\\ \left(ab+\frac{a}{c}+1+\frac{1}{bc}\right)\left( c+\frac{1}{a} \right)&=&\left(1+\frac{1}{b}+\frac{1}{a}+\frac{1}{ab} \right)\left( 1+\frac{1}{c}\right)\\~\\ abc+b+a+\frac{1}{c}+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{abc}&=& 1+\frac{1}{c}+\frac{1}{b}+\frac{1}{bc}+\frac{1}{a}+\frac{1}{ac}+\frac{1}{ab}+\frac{1}{abc}\\~\\ abc+b+a+{\color{gray}\frac{1}{c}}+c+{\color{gray}\frac{1}{a}}+{\color{gray}\frac{1}{b}}+{\color{gray}\frac{1}{abc}}&=& 1+{\color{gray}\frac{1}{c}}+{\color{gray}\frac{1}{b}}+\frac{1}{bc}+{\color{gray}\frac{1}{a}}+\frac{1}{ac}+\frac{1}{ab}+{\color{gray}\frac{1}{abc}}\\~\\ abc+b+a+c&=&1+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}\\~\\ abc+a+b+c&=&1+\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}\\~\\ abc+a+b+c&=&1+\frac{a+b+c}{abc} \end{array}$$

At this point, let's substitute   x   in for   a + b + c   and   13   in for   abc    so we want to solve this for  x:

$$\begin{array}{ccc} 13+x&=&1+\frac{x}{13}\\~\\ 169+13x&=&13+x\\~\\ 13x-x&=&13-169\\~\\ 12x&=&-156\\~\\ x&=&-13 \end{array}$$_

Oct 6, 2019
#2
+2547
+2

Dang thats a lot of work! But a good problem solved.

CalculatorUser  Oct 6, 2019
#3
+1

THANK YOU SO MUCH :)

Oct 7, 2019