2024/053) Given $ab + bc + ca = abc$ find the value of $\frac{(b+c)}{bc(a-1)} + \frac{(c+a)}{ca(b-1)} + \frac{(a+b)}{ab(c-1)}$

 We are given

$ab+bc+ca = abc$

Hence $bc(a-1) = abc - bc = ab + ac = a(b+c)$

Hence

 $\frac{(b+c)}{bc(a-1)} =  \frac{(b+c)}{a(b+c)} = \frac{1}{a}$

or 

 $\frac{(b+c)}{bc(a-1)} = \frac{1}{a}\cdots(1)$

 similarly

 $\frac{c+a)}{ca(b-1)} = \frac{1}{b}\cdots(2)$

and

 $\frac{(a+b)}{ab(c-1)} = \frac{1}{c}\cdots(3)$

Adding (1),(2),(3) we get

 $\frac{(b+c)}{bc(a-1)} + \frac{(c+a)}{ca(b-1)} + \frac{(a+b)}{ab(c-1)}$

$=\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

$=\frac{bc+ca+ab}{abc} = \frac{abc}{abc}=1$