\({1 \over x^2}=x^{-2}\)

\({a^3 \over a^{-7}}=a^{3--7}=a^{10}\)

\(x^3⋅x^{-5}=x^{3+-5}=x^{-2}\)

\((p^5)^{-7}=p^{5⋅-7}=p^{-35}\)

\(a^4⋅{1 \over a^5}=a^4⋅a^{-5}=a^{4+-5}=a^{-1}\)

\({({1 \over a^6}) \over a^5}={a^{-6} \over a^5}=a^{-6-5}=a^{-11}\)

\({a^0 \over a^6}=a^{0-6}=a^{-6}\)

\({9 \over p^2}\)

\({3a \over 5a^5}={3 \over 5}⋅{a \over a^5}={3 \over 5}⋅a^{1-5}={3 \over 5}a^{-4}\)

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

\({7p^5q^2 \over 5pq^6}={7 \over 5}⋅{p^5 \over p^1}⋅{q^2 \over q^6}={7 \over 5}⋅p^{5-1}⋅p^{2-6}=1\frac{2}{5}p^4q^{-4}\)

\(x^2⋅\sqrt[3]{x}=x^2⋅x^{\frac{1}{3}}=x^{2+\frac{1}{3}}=x^{2\frac{1}{3}}\)

\(a^9⋅\sqrt[5]{a^4}=a^9⋅a^{\frac{4}{5}}=a^{9+\frac{4}{5}}=a^{9\frac{4}{5}}\)

\({a^6 \over \sqrt[6]{a^5}}={a^6 \over a^{\frac{5}{6}}}=a^{6-\frac{5}{6}}=a^{5\frac{1}{6}}\)

\({1 \over x^7}⋅\sqrt[8]{x^5}=x^{-7}⋅x^{\frac{5}{8}}=x^{-7+\frac{5}{8}}=x^{-6\frac{3}{8}}\)

\({\sqrt[9]{a^8} \over \sqrt[7]{a^4}}={a^{\frac{8}{9}} \over a^{\frac{4}{7}}}=a^{\frac{8}{9}-\frac{4}{7}}=a^{\frac{20}{63}}\)

\(\sqrt[5]{{1 \over a^2}}=\sqrt[5]{a^{-2}}=a^{-\frac{2}{5}}\)

\(\sqrt{p^4}=p^{\frac{4}{2}}=p^2\)

\({x^8 \over x^4⋅\sqrt[9]{x^2}}={x^8 \over x^4⋅x^{\frac{2}{9}}}={x^8 \over x^{4\frac{2}{9}}}=x^{8-4\frac{2}{9}}=x^{3\frac{7}{9}}\)

\({5b^8 \over 6a^7}\)

\((4x)^{-5}=4^{-5}⋅x^{-5}={1 \over 4^5}⋅{1 \over x^5}={1 \over 1\,024x^5}\)

\(({1 \over 4}p)^{-2}=(4^{-1}⋅p)^{-2}=(4^{-1})^{-2}⋅p^{-2}=4^2⋅p^{-2}={16 \over p^2}\)

\(6x^{8\frac{3}{7}}=6⋅x^8⋅x^{\frac{3}{7}}=6x^8⋅\sqrt[7]{x^3}\)

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