If a - b = 1 and a³ - b³ = 61, then the value of ab will be?

Solution (By Examveda Team)

$$\eqalign{ & {a^3} - {b^3} = 61 \cr & \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = 61 \cr & a - b = 1{\text{ }}\left( {{\text{ }}Given} \right) \cr & 1 \times {a^2} + ab + {b^2} = 61\,......(i) \cr & {\text{Now,}}a - b = 1 \cr & {\text{On squaring both sides}} \cr & {a^2} + {b^2} - 2ab = 1 \cr & {a^2} + {b^2} + ab - 3ab = 1 \cr & {\text{From equation (i)}} \cr & \Rightarrow 61 - 3ab = 1 \cr & \Rightarrow 3ab = 60 \cr & \Rightarrow ab = \frac{{60}}{3} \cr & \Rightarrow ab = 20 \cr} $$