If $$a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$ and $$b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$, then the value of $$\frac{a^2}{b}+\frac{b^2}{a}$$ is:

Given : $$a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$

Rationalizing the denominator, we get :

=> $$a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{(\sqrt3-\sqrt2)}{(\sqrt3-\sqrt2)}$$

=> $$a=\frac{(\sqrt3-\sqrt2)^2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}$$

=> $$a=\frac{3+2-2(\sqrt3)(\sqrt2)}{(3-2)}$$

=> $$a=5-2\sqrt6$$

Similarly, $$b=5+2\sqrt6$$

To find : $$a^{2}+b^{2}$$

= $$(5-2\sqrt6)^2+(5+2\sqrt6)^2$$

= $$(25+24-20\sqrt6)+(25+24+20\sqrt6)$$

= $$49+49=98$$

=> Ans - (A)