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

Answer: Option B

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