Question
If a - b = 1 and a3 - b3 = 61, then the value of ab will be?
Answer: Option B
$$\eqalign{
& {a^3} - {b^3} = 61 \cr
& \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = 61 \cr
& a - b = 1{\text{ }}\left( {{\text{ }}Given} \right) \cr
& 1 \times {a^2} + ab + {b^2} = 61\,......(i) \cr
& {\text{Now,}}a - b = 1 \cr
& {\text{On squaring both sides}} \cr
& {a^2} + {b^2} - 2ab = 1 \cr
& {a^2} + {b^2} + ab - 3ab = 1 \cr
& {\text{From equation (i)}} \cr
& \Rightarrow 61 - 3ab = 1 \cr
& \Rightarrow 3ab = 60 \cr
& \Rightarrow ab = \frac{{60}}{3} \cr
& \Rightarrow ab = 20 \cr} $$
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$$\eqalign{
& {a^3} - {b^3} = 61 \cr
& \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = 61 \cr
& a - b = 1{\text{ }}\left( {{\text{ }}Given} \right) \cr
& 1 \times {a^2} + ab + {b^2} = 61\,......(i) \cr
& {\text{Now,}}a - b = 1 \cr
& {\text{On squaring both sides}} \cr
& {a^2} + {b^2} - 2ab = 1 \cr
& {a^2} + {b^2} + ab - 3ab = 1 \cr
& {\text{From equation (i)}} \cr
& \Rightarrow 61 - 3ab = 1 \cr
& \Rightarrow 3ab = 60 \cr
& \Rightarrow ab = \frac{{60}}{3} \cr
& \Rightarrow ab = 20 \cr} $$
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