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

1(a - b)
2a⁴ + b⁴
3a(a² - ab + b²)
4a³ - a²b² + b²

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let the fourth proportional to \\ (a^{2} - b^{2}), (a^{2} - a b), (a^{3} + b^{3}) be x . \\ Then, \\ = (a^{2} - b^{2}) : (a^{2} - a b) :: (a^{3} + b^{3}) : x \\ \Rightarrow (a^{2} - b^{2}) x = (a^{3} + b^{3}) (a^{2} - a b) \\ \Rightarrow x = \frac{(a^{3} + b^{3}) (a^{2} - a b)}{(a^{2} - b^{2})} \\ = \frac{(a + b) (a^{2} - a b + b^{2}) a (a - b)}{(a - b) (a + b)} \\ \Rightarrow x = a (a^{2} - a b + b) . \end{aligned}

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