If $$x=\sqrt{a}+\frac{1}{\sqrt{a}},y=\sqrt{a}-\frac{1}{\sqrt{a}}, (a>o)$$ the the value of $$x^{4}+y^4-2x^2y^2$$ is 

Given : $$x=\sqrt{a}+\frac{1}{\sqrt{a}}$$

Squaring both sides

=> $$(x)^2=(\sqrt{a}+\frac{1}{\sqrt{a}})^2$$

=> $$x^2=(\sqrt{a})^2+(\frac{1}{\sqrt{a}})^2+2.\sqrt{a}.\frac{1}{\sqrt{a}}$$

=> $$x^2=a+\frac{1}{a}+2$$ -------------(i)

Similarly, $$y^2=a+\frac{1}{a}-2$$ ---------(ii)

To find : $$x^{4}+y^4-2x^2y^2$$

= $$(x^2-y^2)^2$$

Substituting values from equations (i) and (ii), we get :

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

= $$(2+2)^2=(4)^2=16$$

=> Ans - (A)