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

Answer: Option B

$$\begin{array}{rl}& \because a=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ & =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ & =\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}\\ & =\frac{3+2+2\sqrt{6}}{3-2}\\ & =5+2\sqrt{6}\\ & \because b=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ & \text{ = }\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ & \text{ = }\frac{{\left(\sqrt{3}-\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}\\ & =\frac{3+2-2\sqrt{6}}{3-2}\\ & =5-2\sqrt{6}\\ & \therefore a^2+b^2\\ & ={\left(5+2\sqrt{6}\right)}^2+{\left(5-2\sqrt{6}\right)}^2\\ & =2\left[{\left(5\right)}^2+{\left(2\sqrt{6}\right)}^2\right]\\ & =2\left(25+24\right)\\ & =2\times 49\\ & =98\end{array}$$