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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 13, 2021
 #1
avatar+26115 
+2

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 \(a+b+c\).

 

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

 

laugh

 Jun 14, 2021
edited by heureka  Jun 14, 2021
 #2
avatar+2106 
+1

Nice. :))

 

=^._.^=

catmg  Jun 14, 2021

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