The trick is to exploit the difference of squares formula,
[tex]a^2-b^2=(a-b)(a+b)[/tex]
Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:
[tex](\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2[/tex]
Whatever you do to the denominator, you have to do to the numerator too. So
[tex]\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2[/tex]
Expand the numerator:
[tex](\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2[/tex]
[tex](\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6[/tex]
[tex](\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6[/tex]
[tex](\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)[/tex]
[tex](\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)[/tex]
[tex](\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}[/tex]
So we have
[tex]\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2[/tex]
But √12 = √(3•4) = 2√3, so
[tex]\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}[/tex]
