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Question 1

Question 2

$\frac{a^{-5/4}-a^{-2}}{a^{-7/4}-a^{-2}} -\frac{a^{-1/2}-a^{-3/2}}{a^{-1}-a^{-3/2}}= \frac{a^{-5/4}(1-a^{-3/4)}}{a^{-5/4}(a^{-2/4}-a^{-3/4})} -\frac{a^{-1/2}(1-a^{-1})}{a^{-1/2}(a^{-1/2}-a^{-1})}=$

$= \frac{1-a^{-3/4}}{a^{-2/4}-a^{-3/4}} -\frac{1-a^{-1}}{a^{-1/2}-a^{-1}}= \frac{1-a^{-3/4}}{a^{-2/4}(1-a^{-1/4})} -\frac{1-a^{-1}}{a^{-1/2}(1-a^{-1/2})}=$

$=\frac{1-a^{-3/4}}{a^{-1/2}(1-a^{-1/4})} -\frac{1-a^{-1}}{a^{-1/2}(1-a^{-1/4})(1+a^{-1/4})}=$

$=\frac{(1-a^{-3/4})(1+a^{-1/4})-(1-a^{-1})}{a^{-1/2}(1-a^{-1/4})(1+a^{-1/4})}=\frac{(1^3-(a^{-1/4})^3)(1+a^{-1/4})-(1^2-(a^{-1/2})^2)}{a^{-1/2}(1-a^{-1/4})(1+a^{-1/4})}=$

$=\frac{(1-a^{-1/4})(1+a^{-1/4}+a^{-1/2})(1+a^{-1/4})-(1+a^{-1/2})(1-a^{-1/2})}{a^{-1/2}(1-a^{-1/4})(1+a^{-1/4})}=$

$=\frac{(1-a^{-1/2})(1+a^{-1/4}+a^{-1/2})-(1+a^{-1/2})(1-a^{-1/2})}{a^{-1/2}(1-a^{-1/2})}=$

$=\frac{(1-a^{-1/2})[(1+a^{-1/4}+a^{-1/2})-(1+a^{-1/2})]}{a^{-1/2}(1-a^{-1/2})}= \frac{1+a^{-1/4}+a^{-1/2}-1-a^{-1/2}}{a^{-1/2}}=$

$=\frac{a^{-1/4}}{a^{-1/2}}=a^{-1/4-(-1/2)}=a^{1/4}= \sqrt[4]{a}$

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$(a^{1/3}-x^{1/3})^{-1}*(a-x)= \frac{1}{a^{1/3}-x^{1/3}} *[(a^{1/3})^3-(x^{1/3})^3]=$

$=\frac{(a^{1/3})^3-(x^{1/3})^3}{a^{1/3}-x^{1/3}} = \frac{(a^{1/3}-x^{1/3})(a^{2/3}+a^{1/3}x^{1/3}+x^{2/3})}{a^{1/3}-x^{1/3}} =$

$=a^{2/3}+a^{1/3}x^{1/3}+x^{2/3}= \sqrt[3]{a^2} + \sqrt[3]{ax} + \sqrt[3]{x^2}$