Given that $A = (\sqrt{2008}+\sqrt{2009}),$ $B = (-\sqrt{2008}-\sqrt{2009}),$ $C = (\sqrt{2008}-\sqrt{200}),$ and $D = (\sqrt{200}-\sqrt{2008}),$ find $ABCD.$

Answer 1

We see that $$(A-B)^2=((\sqrt{2008}+\sqrt{2009})-(-\sqrt{2008}-\sqrt{2009}))^2=(2\sqrt{2008})^2=4(2008).$$We note that the square root of the $2008$s in the answer choices are $2 \sqrt{4 \cdot 502} = 4 \sqrt{502}.$ The product of the last two terms in $(C-D)^2$ is $$((\sqrt{2008}-\sqrt{200})-(\sqrt{200}-\sqrt{2008}))(2 \sqrt{2008}) = -4(200-2008) = 4(2008).$$Therefore, $(A-B)^2=(C-D)^2$ so $ABCD = (A-B)(C-D) = \left(4 \sqrt{2008}\right) \left(-4 \sqrt{502}\right) = \boxed{-4(2008)}.$