For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ . I am trying to find $x+y$ given that
$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$
It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side 
$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$
and see if a term $x+y$ comes out, but it looks hopeless at this point. 

$$
\begin{align}
x+\sqrt{x^2+3}
&=\frac3{y+\sqrt{y^2+3}}\\
&=\frac3{y+\sqrt{y^2+3}}\frac{y-\sqrt{y^2+3}}{y-\sqrt{y^2+3}}\\
&=\frac3{-3}\left(y-\sqrt{y^2+3}\right)\\
&=-y+\sqrt{y^2+3}\tag{1}
\end{align}
$$
Similarly
$$
y+\sqrt{y^2+3}=-x+\sqrt{x^2+3}\tag{2}
$$
Add $(1)$ and $(2)$ and cancel the radicals.
