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# Algebra

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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 $$a +b+c$$.

Mar 6, 2020

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Multiplying out and simplifying the left hand side:  (a + 1/b)(b + 1/c)(c + 1/a)

gives you:  abc + (a + b + c) + (1/a + 1/b + 1/c) + 1/(abc)

Multiplying out and simplifying the right hand side:  (1 + 1/a)(1 + 1/b)(1 + 1/c)

gives you:  1 + (1/a + 1/b + 1/c) + ( 1/(ab) + 1/(ac) + 1/(bc) ) + 1/(abc)

Setting these two equal to each other and subtracting both (1/a + 1/b + 1/c) and 1/(abc) from both sides,

leaves you with:  abc + (a + b + c)  =  1 + 1/(ab) + 1/(ac) + 1/(bc)

Rewrite 1/(ab) + 1/(ac) + 1/(bc)  as  (a + b + c)/(abc)

the equation becomes:   abc + (a + b + c)  =  1 + (a + b + c)/(abc)

Since  abc = 13:               13 + (a + b + c)  =  1 + (a + b + c)/13

Subtract 1:                        12 + (a + b + c)  =  (a + b + c)/13

Multiply by 13:            156 + 13(a + b + c)  =  (a + b + c)

Simplify:                     156 + 12(a + b + c)  =  0

12(a + b + c)  =  -156

a + b + c  =  -13

Mar 6, 2020