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July 2022 1 10 Report

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[tex](\frac{2}{a^{\frac{2}{3}}-4}}+\frac{2}{2-a^{\frac{1}{3}}}-\frac{1}{2+a^{\frac{1}{3}}}): ( \frac{4-a^{\frac{2}{3}}}{3})^{-1}[/tex]

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

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

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