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

\({p^2 \over p^{-6}}=p^{2--6}=p^8\)

\(a^2⋅a^{-7}=a^{2+-7}=a^{-5}\)

\((a^9)^{-8}=a^{9⋅-8}=a^{-72}\)

\(x^3⋅{1 \over x^9}=x^3⋅x^{-9}=x^{3+-9}=x^{-6}\)

\({({1 \over x^4}) \over x^3}={x^{-4} \over x^3}=x^{-4-3}=x^{-7}\)

\({a^7 \over a^0}=a^{7-0}=a^7\)

\({5 \over a^9}\)

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

\({x^6 \over ({1 \over x^9})}={x^6 \over x^{-9}}=x^{6--9}=x^{15}\)

\({8a^6b^2 \over 7a^5b^4}={8 \over 7}⋅{a^6 \over a^5}⋅{b^2 \over b^4}={8 \over 7}⋅a^{6-5}⋅a^{2-4}=1\frac{1}{7}ab^{-2}\)

\(p^4⋅\sqrt[8]{p}=p^4⋅p^{\frac{1}{8}}=p^{4+\frac{1}{8}}=p^{4\frac{1}{8}}\)

\(x^8⋅\sqrt[7]{x^3}=x^8⋅x^{\frac{3}{7}}=x^{8+\frac{3}{7}}=x^{8\frac{3}{7}}\)

\({x^3 \over \sqrt[9]{x^4}}={x^3 \over x^{\frac{4}{9}}}=x^{3-\frac{4}{9}}=x^{2\frac{5}{9}}\)

\({1 \over a^4}⋅\sqrt[9]{a^7}=a^{-4}⋅a^{\frac{7}{9}}=a^{-4+\frac{7}{9}}=a^{-3\frac{2}{9}}\)

\({\sqrt[8]{a^7} \over \sqrt[5]{a^3}}={a^{\frac{7}{8}} \over a^{\frac{3}{5}}}=a^{\frac{7}{8}-\frac{3}{5}}=a^{\frac{11}{40}}\)

\(\sqrt[8]{{1 \over p^7}}=\sqrt[8]{p^{-7}}=p^{-\frac{7}{8}}\)

\(\sqrt[4]{x^{16}}=x^{\frac{16}{4}}=x^4\)

\({x^9 \over x^6⋅\sqrt[9]{x^4}}={x^9 \over x^6⋅x^{\frac{4}{9}}}={x^9 \over x^{6\frac{4}{9}}}=x^{9-6\frac{4}{9}}=x^{2\frac{5}{9}}\)

\({3y^3 \over 5x^4}\)

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

\(({1 \over 4}a)^{-3}=(4^{-1}⋅a)^{-3}=(4^{-1})^{-3}⋅a^{-3}=4^3⋅a^{-3}={64 \over a^3}\)

\(6a^{5\frac{3}{4}}=6⋅a^5⋅a^{\frac{3}{4}}=6a^5⋅\sqrt[4]{a^3}\)

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