If (x - a)(x - b) = 1 and a - b + 5 = 0, then the value of $${\left( {x - a} \right)^3}$$   - $$\frac{1}{{{{\left( {x - a} \right)}^3}}}\,{\text{}}$$  is?

Solution (By Examveda Team)

$$\eqalign{ & \left( {x - a} \right)\left( {x - b} \right) = 1{\text{ }} \cr & \Rightarrow \left( {x - a} \right) = \frac{1}{{\left( {x - b} \right)}} \cr & {\text{ }}a - b + 5 = 0 \cr & \Rightarrow {\text{ }}a - b = - 5{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{Add and subtract }}x \cr & \Rightarrow a - b + x - x = - 5 \cr & \Rightarrow \left( {a - x} \right) + \left( {x - b} \right) = - 5 \cr & \Rightarrow \left( {x - b} \right) - \left( {x - a} \right) = - 5 \cr & \Rightarrow \left( {x - a} \right) - \left( {x - b} \right) = + 5 \cr & \Rightarrow \left( {x - a} \right) - \frac{1}{{\left( {x - a} \right)}} = + 5 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} - 3\left( {x - a} \right)\frac{1}{{\left( {x - a} \right)}}\left( {\left( {x - a} \right) - \frac{1}{{\left( {x - a} \right)}}} \right) = {\left( 5 \right)^3} \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} - 3 \times 5 = 125 \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} = 125 + 15 \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} = 140 \cr} $$