If a,b,c are real numbers such that
**
$\frac{ab+1}{b}$
=
$\frac{7}{3}$
**
**
$\frac{bc+1}{c}$
=4
**
**
$\frac{ac+1}{a}$
=1
**

Find a×b×c

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The answer is 1.

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Multiplying all the 3 equations:

$(a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{a})=\frac{28}{3}$

$abc+\frac{1}{abc}+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{28}{3}$

putting the values from the given equations we get:

$abc+\frac{1}{abc}=2$

Solving the equation gives

Hence $abc=1$