Fourth proportional to (a² - b²), (a² - ab), (a³ + b³) is

Solution (By Examveda Team)

$$\eqalign{ & {\text{Let the fourth proportional to}} \cr & \left( {{a^2} - {b^2}} \right),\left( {{a^2} - ab} \right),\left( {{a^3} + {b^3}} \right)\,{\text{be}}\,x \cr & {\text{Then,}} \cr & = \left( {{a^2} - {b^2}} \right):\left( {{a^2} - ab} \right)::\left( {{a^3} + {b^3}} \right):x \cr & \Rightarrow \left( {{a^2} - {b^2}} \right)x = \left( {{a^3} + {b^3}} \right)\left( {{a^2} - ab} \right) \cr & \Rightarrow x = \frac{{\left( {{a^3} + {b^3}} \right)\left( {{a^2} - ab} \right)}}{{\left( {{a^2} - {b^2}} \right)}} \cr & \Rightarrow x = \frac{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)a\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {a + b} \right)}} \cr & \Rightarrow x = a\left( {{a^2} - ab + b} \right) \cr} $$