\({1 \over a^3}=a^{-3}\)

\({4x^3 \over 7x^6}={4 \over 7}⋅{x^3 \over x^6}={4 \over 7}⋅x^{3-6}={4 \over 7}x^{-3}\)

\({a^6 \over a^{-8}}=a^{6--8}=a^{14}\)

\(x^3⋅x^{-6}=x^{3+-6}=x^{-3}\)

\((p^3)^{-7}=p^{3⋅-7}=p^{-21}\)

\(x^4⋅{1 \over x^6}=x^4⋅x^{-6}=x^{4+-6}=x^{-2}\)

\({({1 \over a^9}) \over a^2}={a^{-9} \over a^2}=a^{-9-2}=a^{-11}\)

\({x^4 \over ({1 \over x^9})}={x^4 \over x^{-9}}=x^{4--9}=x^{13}\)

\({6x^5y^2 \over 5x^3y^3}={6 \over 5}⋅{x^5 \over x^3}⋅{y^2 \over y^3}={6 \over 5}⋅x^{5-3}⋅x^{2-3}=1\frac{1}{5}x^2y^{-1}\)

\({a^0 \over a^8}=a^{0-8}=a^{-8}\)

\(x^7⋅\sqrt[5]{x}=x^7⋅x^{\frac{1}{5}}=x^{7+\frac{1}{5}}=x^{7\frac{1}{5}}\)

\(p^8⋅\sqrt[9]{p^2}=p^8⋅p^{\frac{2}{9}}=p^{8+\frac{2}{9}}=p^{8\frac{2}{9}}\)

\({a^2 \over \sqrt[7]{a^4}}={a^2 \over a^{\frac{4}{7}}}=a^{2-\frac{4}{7}}=a^{1\frac{3}{7}}\)

\({1 \over p^5}⋅\sqrt[5]{p^2}=p^{-5}⋅p^{\frac{2}{5}}=p^{-5+\frac{2}{5}}=p^{-4\frac{3}{5}}\)

\({\sqrt[7]{x^4} \over \sqrt[9]{x^4}}={x^{\frac{4}{7}} \over x^{\frac{4}{9}}}=x^{\frac{4}{7}-\frac{4}{9}}=x^{\frac{8}{63}}\)

\(\sqrt[7]{{1 \over a^2}}=\sqrt[7]{a^{-2}}=a^{-\frac{2}{7}}\)

\(\sqrt[3]{x^9}=x^{\frac{9}{3}}=x^3\)

\({x^9 \over x^3⋅\sqrt[8]{x^7}}={x^9 \over x^3⋅x^{\frac{7}{8}}}={x^9 \over x^{3\frac{7}{8}}}=x^{9-3\frac{7}{8}}=x^{5\frac{1}{8}}\)

\({9 \over a^8}\)

\({4q^9 \over 5p^7}\)

\(({1 \over 2}a)^{-5}=(2^{-1}⋅a)^{-5}=(2^{-1})^{-5}⋅a^{-5}=2^5⋅a^{-5}={32 \over a^5}\)

\(4p^{2\frac{5}{7}}=4⋅p^2⋅p^{\frac{5}{7}}=4p^2⋅\sqrt[7]{p^5}\)

\(\frac{1}{3}a^{-\frac{7}{8}}b^{\frac{7}{9}}=\frac{1}{3}⋅{1 \over a^{\frac{7}{8}}}⋅b^{\frac{7}{9}}={1⋅\sqrt[9]{b^7} \over 3⋅\sqrt[8]{a^7}}\)

\((5a)^{-2}=5^{-2}⋅a^{-2}={1 \over 5^2}⋅{1 \over a^2}={1 \over 25a^2}\)