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

\({8 x^{2} \over 9 x^{4}} = {8 \over 9} ⋅ {x^{2} \over x^{4}} = {8 \over 9} ⋅ x^{2 - 4} = {8 \over 9} x^{-2}\)

\({p^{2} \over p^{-4}} = p^{2 - -4} = p^{6}\)

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

\((a^{8})^{-9} = a^{8 ⋅ -9} = a^{-72}\)

\(x^{5} ⋅ {1 \over x^{9}} = x^{5} ⋅ x^{-9} = x^{5 + -9} = x^{-4}\)

\({({1 \over x^{9}}) \over x^{3}} = {x^{-9} \over x^{3}} = x^{-9 - 3} = x^{-12}\)

\({a^{7} \over ({1 \over a^{8}})} = {a^{7} \over a^{-8}} = a^{7 - -8} = a^{15}\)

\({3 x^{4} y^{5} \over 2 x^{3} y^{6}} = {3 \over 2} ⋅ {x^{4} \over x^{3}} ⋅ {y^{5} \over y^{6}} = {3 \over 2} ⋅ x^{4 - 3} ⋅ x^{5 - 6} = 1\frac{1}{2} x y^{-1}\)

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

\(p^{6} ⋅ \sqrt[8]{p} = p^{6} ⋅ p^{\frac{1}{8}} = p^{6 + \frac{1}{8}} = p^{6\frac{1}{8}}\)

\(x^{9} ⋅ \sqrt[9]{x^{7}} = x^{9} ⋅ x^{\frac{7}{9}} = x^{9 + \frac{7}{9}} = x^{9\frac{7}{9}}\)

\({x^{9} \over \sqrt[4]{x^{3}}} = {x^{9} \over x^{\frac{3}{4}}} = x^{9 - \frac{3}{4}} = x^{8\frac{1}{4}}\)

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

\({\sqrt[8]{x^{5}} \over \sqrt[7]{x^{3}}} = {x^{\frac{5}{8}} \over x^{\frac{3}{7}}} = x^{\frac{5}{8} - \frac{3}{7}} = x^{\frac{11}{56}}\)

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

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

\({p^{2} \over p^{5} ⋅ \sqrt[7]{p^{5}}} = {p^{2} \over p^{5} ⋅ p^{\frac{5}{7}}} = {p^{2} \over p^{5\frac{5}{7}}} = p^{2 - 5\frac{5}{7}} = p^{-3\frac{5}{7}}\)

\({6 \over p^{2}}\)

\({2 b^{3} \over 7 a^{7}}\)

\(({1 \over 3} a)^{-2} = (3^{-1} ⋅ a)^{-2} = (3^{-1})^{-2} ⋅ a^{-2} = 3^{2} ⋅ a^{-2} = {9 \over a^{2}}\)

\(6 x^{3\frac{4}{9}} = 6 ⋅ x^{3} ⋅ x^{\frac{4}{9}} = 6 x^{3} ⋅ \sqrt[9]{x^{4}}\)

\(\frac{1}{6} a^{-\frac{2}{7}} b^{\frac{4}{9}} = \frac{1}{6} ⋅ {1 \over a^{\frac{2}{7}}} ⋅ b^{\frac{4}{9}} = {1 ⋅ \sqrt[9]{b^{4}} \over 6 ⋅ \sqrt[7]{a^{2}}}\)

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