If a = 3 - 2 3 + 2 and b = 3 + 2 3 - 2 , then the value of a^2b + b^2a = ?

Answer: Option B

$$\begin{array}{rl}& \text{Given,}\\ & \because a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\text{, }b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ & \text{Find }\frac{a^2}{b}+\frac{b^2}{a}=?\\ & \Rightarrow \frac{a^3+b^3}{ab}=?\\ & \Rightarrow \frac{{\left(a+b\right)}^2-3ab\left(a+b\right)}{ab}=?\\ & \Rightarrow a+b=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ & \Rightarrow \frac{{\left(\sqrt{3}-\sqrt{2}\right)}^2+{\left(\sqrt{3}+\sqrt{2}\right)}^2}{{\sqrt{3}}^2-{\sqrt{2}}^2}\\ & \Rightarrow \frac{2\left({\sqrt{3}}^2+{\sqrt{2}}^2\right)}{3-2}\\ & \Rightarrow \frac{2\times \left(5\right)}{1}\\ & \Rightarrow a+b=10\\ & \text{Again, }\\ & \Rightarrow a\times b=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ & \Rightarrow ab=1\\ & \Rightarrow \frac{{\left(a+b\right)}^3-3ab\left(a+b\right)}{ab}\\ & \Rightarrow \frac{{10}^3-3\times 1\times 10}{1}\\ & \Rightarrow 1000-30\\ & \Rightarrow 970\end{array}$$