Answer: C
Let the fourth proportional be \(p\) \(\therefore \frac{(x^2 - xy + y^2)}{(x^3 + y^3)} = \frac{(x - y)}{p}\) ⇒ \(p(x^2 - xy + y^2) = (x - y)(x^3 + y^3)\) ⇒ \(p = \frac{(x - y)(x^3 + y^3)}{(x^2 - xy + y^2)}\) \(= \frac{(x - y)(x + y)\cancel{(x^2 - xy + y^2)}}{\cancel{(x^2 - xy + y^2)}}\) \(= x^2 - y^2\)
Let the fourth proportional be \(p\) \(\therefore \frac{(x^2 - xy + y^2)}{(x^3 + y^3)} = \frac{(x - y)}{p}\) ⇒ \(p(x^2 - xy + y^2) = (x - y)(x^3 + y^3)\) ⇒ \(p = \frac{(x - y)(x^3 + y^3)}{(x^2 - xy + y^2)}\) \(= \frac{(x - y)(x + y)\cancel{(x^2 - xy + y^2)}}{\cancel{(x^2 - xy + y^2)}}\) \(= x^2 - y^2\)