If $$a - \frac{1}{a} = b,\,b - \frac{1}{b} = c$$     and $$c - \frac{1}{c} = a,$$   then what is the value $$\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}} = ?$$

Solution (By Examveda Team)

$$\eqalign{ & a - \frac{1}{a} = b\,........\left( {\text{i}} \right) \cr & b - \frac{1}{b} = c\,........\left( {{\text{ii}}} \right) \cr & c - \frac{1}{c} = a\,........\left( {{\text{iii}}} \right) \cr & {\text{Add Equation }}\left( {\text{i}} \right),\,\left( {{\text{ii}}} \right){\text{ and}}\left( {{\text{iii}}} \right) \cr & a + b + c - \left[ {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right] = a + b + c \cr & \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\,........\left( {{\text{iv}}} \right) \cr & \Rightarrow a - \frac{1}{a} = b \cr & {\text{Squaring both sides}} \cr & {a^2} + \frac{1}{{{a^2}}} - 2 = {b^2}\,........\left( {\text{v}} \right) \cr & \Rightarrow b - \frac{1}{b} = c \cr & {b^2} + \frac{1}{{{b^2}}} - 2 = {c^2}\,........\left( {{\text{vi}}} \right) \cr & \Rightarrow c + \frac{1}{c} = a \cr & {c^2} + \frac{1}{{{c^2}}} - 2 = {a^2}\,........\left( {{\text{vii}}} \right) \cr & {\text{Add Equation }}\left( {\text{v}} \right),\,\left( {{\text{vi}}} \right){\text{ and}}\left( {{\text{vii}}} \right) \cr & {a^2} + {b^2} + {c^2} + \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} - 2 - 2 - 2 = {a^2} + {b^2} + {c^2} \cr & \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 6\,........\left( {{\text{viii}}} \right) \cr & \Rightarrow \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \cr & = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + \frac{2}{{ab}} + \frac{2}{{bc}} + \frac{2}{{ca}} \cr & {0^2} = 6 + 2\left[ {\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}}} \right] \cr & \frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}} = \frac{{ - 6}}{2} = - 3 \cr} $$