$\Bigg(\dfrac{a\sqrt[]{a}-1}{a-\sqrt[]{a}}-\dfrac{a\sqrt[]{a}+1}{a+\sqrt[]{a}}\Bigg):\dfrac{a+2}{a-2}$ $=\Bigg[\dfrac{(\sqrt[]{a}-1)(a+\sqrt[]{a}+1)}{\sqrt[]{a}(\sqrt[]{a}-1)}-\dfrac{(\sqrt[]{a}+1)(a-\sqrt[]{a}+1)}{\sqrt[]{a}(\sqrt[]{a}+1)}\Bigg].\dfrac{a-2}{a+2}$ $=\dfrac{a+\sqrt[]{a}+1-(a-\sqrt[]{a}+1)}{\sqrt[]{a}}.\dfrac{a-2}{a+2}$ $=\dfrac{2a-4}{a+2}$ Reply
$\Bigg(\dfrac{a\sqrt[]{a}-1}{a-\sqrt[]{a}}-\dfrac{a\sqrt[]{a}+1}{a+\sqrt[]{a}}\Bigg):\dfrac{a+2}{a-2}$
$=\Bigg[\dfrac{(\sqrt[]{a}-1)(a+\sqrt[]{a}+1)}{\sqrt[]{a}(\sqrt[]{a}-1)}-\dfrac{(\sqrt[]{a}+1)(a-\sqrt[]{a}+1)}{\sqrt[]{a}(\sqrt[]{a}+1)}\Bigg].\dfrac{a-2}{a+2}$
$=\dfrac{a+\sqrt[]{a}+1-(a-\sqrt[]{a}+1)}{\sqrt[]{a}}.\dfrac{a-2}{a+2}$
$=\dfrac{2a-4}{a+2}$