Question

If $a.b$ and $c$ are positive numbers satisfying $a+\frac{1}{b}=4,b+\frac{1}{c}=1$ and $c+\frac{1}{a}=\frac{7}{3}$ then the value of $6$ $abc$ is

Answer 1

$\begin{array}{c}Wehave\\ a+\frac{1}{b}=4\Rightarrow a=4-\frac{1}{b}\\ b+\frac{1}{c}=1\Rightarrow b=1-\frac{1}{c}\\ c+\frac{1}{a}=\frac{7}{3}\Rightarrow c=\frac{7}{3}-\frac{1}{a}\\ Now,\\ 6abc=6\times \left(4-\frac{1}{b}\right)\times \left(1-\frac{1}{c}\right)\times \left(\frac{7}{3}-\frac{1}{a}\right)\\ =6\left(\frac{4b-1}{b}\right)\left(\frac{c-1}{c}\right)\left(\frac{7a-3}{a}\right)\\ =\left(\frac{24b-6}{b}\right)\left(\frac{7ac-3c-7a+3}{ca}\right)\\ =\frac{\left(24b-6\right)\left(7ac-7a-3c+3\right)}{abc}\end{array}$