If $$ a =\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$ and $$b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$ then $$\frac{a^2}{b}+\frac{b^2}{a}$$

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

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

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

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

Squaring both sides, we get : $$a^2=49-20\sqrt6$$

Similarly, $$b=5+2\sqrt6$$ and $$b^2=49+20\sqrt6$$

To find : $$\frac{a^2}{b}+\frac{b^2}{a}$$

= $$\frac{a^3+b^3}{ab}=\frac{(a+b)(a^2+b^2-ab)}{ab}$$

= $$\frac{[(5-2\sqrt6)+(5+2\sqrt6)][(49-20\sqrt6)+(49+20\sqrt6)-(5-2\sqrt6)(5+2\sqrt6)]}{(5-2\sqrt6)(5+2\sqrt6)}$$

= $$\frac{10[49+49-(25-24)]}{25-24}$$

= $$10\times97=970$$

=> Ans - (B)