We have
[tex]\frac{\sqrt[]{8}}{\sqrt[]{2}-2}[/tex][tex]\frac{\sqrt[]{4\cdot2}}{\sqrt[]{2}-2}[/tex][tex]\frac{\sqrt[]{4}\cdot\sqrt[]{2}}{\sqrt[]{2}-2}[/tex][tex]\frac{2\sqrt[]{2}}{\sqrt[]{2}-2}[/tex]We rationalize the denominator
[tex]\frac{\sqrt[]{2}+2}{\sqrt[]{2}+2}\cdot\frac{2\sqrt[]{2}}{\sqrt[]{2}-2}[/tex]then we simplify
[tex]\frac{4+4\sqrt[]{2}}{-2}[/tex]Finally, we obtain
[tex]-2-2\sqrt[]{2}[/tex]