If x is a rational number and $$\frac{{{{\left( {x + 1} \right)}^3} - {{\left( {x - 1} \right)}^3}}}{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2}}}$$     = 2, then the sum of numerator and denominator of x is?

Solution (By Examveda Team)

$$\eqalign{ & \frac{{{{\left( {x + 1} \right)}^3} - {{\left( {x - 1} \right)}^3}}}{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2}}} = 2 \cr & {{\text{A}}^3} - {{\text{B}}^3} = \left( {{\text{A}} - {\text{B}}} \right)\left( {{{\text{A}}^2} + {\text{AB}} + {{\text{B}}^2}} \right) \cr & {{\text{A}}^2} - {{\text{B}}^2} = \left( {{\text{A}} - {\text{B}}} \right)\left( {{\text{A}} + {\text{B}}} \right) \cr} $$
$$ \Rightarrow \frac{{\left( {x + 1 - x + 1} \right)\left\{ {{{\left( {x + 1} \right)}^2} + \left( {x - 1} \right)\left( {x + 1} \right) + {{\left( {x - 1} \right)}^2}} \right\}}}{{\left( {x + 1 - x + 1} \right)\left( {x + 1 + x - 1} \right)}} = 2$$

$$\eqalign{ & \Rightarrow \frac{{\left( {{x^2} + 1 + 2x + {x^2} - 1 + {x^2} + 1 - 2x} \right)}}{{\left( {2x} \right)}} = 2 \cr & \Rightarrow \frac{{3{x^2} + 1}}{{2x}} = 2 \cr & \Rightarrow 3{x^2} + 1 = 4x \cr & \Rightarrow 3{x^2} - 4x + 1 = 0 \cr & \Rightarrow 3{x^2} - 3x - x + 1 = 0 \cr & \Rightarrow 3x\left( {x - 1} \right) - 1\left( {x - 1} \right) = 0 \cr & \Rightarrow \left( {3x - 1} \right)\left( {x - 1} \right) = 0 \cr & 3x - 1 = 0 \cr & \Leftrightarrow x = \frac{1}{3} \cr & x - 1 = 0 \cr & \text{for } x = 1 = \frac{1}{1} \cr & {\text{By adding numinator and denominator }} \cr & 1 + 1 = 2 \cr & {\text{No option is satisfied}} \cr & \therefore x = \frac{1}{3} \cr & \Rightarrow x = 1 + 3 \cr & \Rightarrow x = 4 \cr} $$