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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.

 Jun 23, 2020
 #1
avatar+25225 
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If

\(abc = 13 \)

and
\(\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)\left(c+\dfrac{1}{a}\right)=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\),
find \(a + b + c\).

 

\(\small{ \begin{array}{|rcll|} \hline \mathbf{\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)\left(c+\dfrac{1}{a}\right) } &=& \mathbf{\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)} \\\\ \left(\dfrac{ab+1}{b}\right)\left(\dfrac{bc+1}{c}\right)\left(\dfrac{ca+1}{a}\right) &=& \left(\dfrac{1+a}{a}\right)\left(\dfrac{1+b}{b}\right)\left(\dfrac{1+c}{c}\right) \\\\ \dfrac{(ab+1)(bc+1)(ca+1)}{abc} &=& \dfrac{(1+a)(1+b)(1+c)}{abc} \\\\ \dfrac{(ab+1)(bc+1)(ca+1)}{abc}-\dfrac{(1+a)(1+b)(1+c)}{abc} &=& 0 \\\\ \dfrac{(ab+1)(bc+1)(ca+1)-(1+a)(1+b)(1+c)}{abc} &=& 0 \\\\ (ab+1)(bc+1)(ca+1)-(1+a)(1+b)(1+c) &=& 0 \\\\ (ab^2c+ab+bc+1)(ca+1)-(ab+a+b+1)(c+1) &=& 0 \\ a^2b^2c^2+a^2bc+abc^2+ca+ab^2c+ab+bc+1-(abc+ab+ac+a+bc+b+c+1) &=& 0 \\ a^2b^2c^2+a^2bc+abc^2+ca+ab^2c+ab+bc+1-abc-ab-ac-a-bc-b-c-1 &=& 0 \\ a^2b^2c^2+a^2bc+abc^2+ab^2c-abc-a-b-c &=& 0 \\ abc(abc+a+b+c)-(abc+a+b+c) &=& 0 \\ (abc-1)(abc+a+b+c) &=& 0 \quad | \quad abc= 13 \\ (13-1)(13+a+b+c) &=& 0 \\ 12(13+a+b+c) &=& 0 \\ 13+a+b+c &=& 0 \\ \mathbf{a+b+c} &=& \mathbf{-13} \\ \hline \end{array} } \)

 

laugh

 
 Jun 24, 2020
edited by heureka  Jun 24, 2020

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