Determine the value of $$\frac{a^2+b^2}{a^3+b^3}$$  $$a=2+\sqrt{3}$$ and $$b=2-\sqrt{3}$$

Given : $$a=2+\sqrt{3}$$

=> $$\frac{1}{a}=\frac{1}{2+\sqrt{3}}$$

=> $$\frac{1}{a}=\frac{1}{2+\sqrt{3}}\times(\frac{2-\sqrt3}{2-\sqrt3})$$

=> $$\frac{1}{a}=\frac{2-\sqrt3}{4-3}=2-\sqrt3=b$$

Thus, $$b=\frac{1}{a}$$

=> $$a+\frac{1}{a}=(2+\sqrt3)+(2-\sqrt3)=4$$ -------------(i)

To find : $$\frac{a^2+b^2}{a^3+b^3}=(a^2+\frac{1}{a^2})\div(a^3+\frac{1}{a^3})$$

= $$[(a+\frac{1}{a})^2-2(a)(\frac{1}{a})]\div[(a+\frac{1}{a})^3-3(a)(\frac{1}{a})(a+\frac{1}{a})]$$

Using equation (i), we get :

= $$[(4)^2-2]\div[(4)^3-3(4)]$$

= $$\frac{(16-2)}{(64-12)}=\frac{14}{52}\approx0.27$$

=> Ans - (A)