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

\({7 a \over 9 a^{5}} = {7 \over 9} ⋅ {a \over a^{5}} = {7 \over 9} ⋅ a^{1 - 5} = {7 \over 9} a^{-4}\)

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

\(a^{2} ⋅ a^{-8} = a^{2 + -8} = a^{-6}\)

\((p^{7})^{-6} = p^{7 ⋅ -6} = p^{-42}\)

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

\({({1 \over p^{5}}) \over p^{2}} = {p^{-5} \over p^{2}} = p^{-5 - 2} = p^{-7}\)

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

\({7 x^{6} y^{3} \over 5 x^{2} y^{6}} = {7 \over 5} ⋅ {x^{6} \over x^{2}} ⋅ {y^{3} \over y^{6}} = {7 \over 5} ⋅ x^{6 - 2} ⋅ x^{3 - 6} = 1\frac{2}{5} x^{4} y^{-3}\)

\({a^{0} \over a^{3}} = a^{0 - 3} = a^{-3}\)

\(x^{6} ⋅ \sqrt[9]{x} = x^{6} ⋅ x^{\frac{1}{9}} = x^{6 + \frac{1}{9}} = x^{6\frac{1}{9}}\)

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

\({x^{2} \over \sqrt[9]{x^{7}}} = {x^{2} \over x^{\frac{7}{9}}} = x^{2 - \frac{7}{9}} = x^{1\frac{2}{9}}\)

\({1 \over a^{4}} ⋅ \sqrt[9]{a^{5}} = a^{-4} ⋅ a^{\frac{5}{9}} = a^{-4 + \frac{5}{9}} = a^{-3\frac{4}{9}}\)

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

\(\sqrt[7]{{1 \over p^{5}}} = \sqrt[7]{p^{-5}} = p^{-\frac{5}{7}}\)

\(\sqrt[3]{x^{6}} = x^{\frac{6}{3}} = x^{2}\)

\({a^{8} \over a^{5} ⋅ \sqrt[3]{a^{2}}} = {a^{8} \over a^{5} ⋅ a^{\frac{2}{3}}} = {a^{8} \over a^{5\frac{2}{3}}} = a^{8 - 5\frac{2}{3}} = a^{2\frac{1}{3}}\)

\({4 \over a^{9}}\)

\({3 y^{7} \over 4 x^{5}}\)

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

\(6 x^{8\frac{7}{9}} = 6 ⋅ x^{8} ⋅ x^{\frac{7}{9}} = 6 x^{8} ⋅ \sqrt[9]{x^{7}}\)

\(\frac{4}{7} a^{-\frac{2}{5}} b^{\frac{1}{6}} = \frac{4}{7} ⋅ {1 \over a^{\frac{2}{5}}} ⋅ b^{\frac{1}{6}} = {4 ⋅ \sqrt[6]{b} \over 7 ⋅ \sqrt[5]{a^{2}}}\)

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