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

Answer: Option C

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