6. \( \sqrt[3]{\frac{x^{2}}{y}} \cdot \sqrt[12]{\frac{27 x^{2}}{16 y}}: \sqrt[6]{\frac{9 x^{5}}{4 y^{2}}} \) 7. \( \sqrt[8]{\frac{x^{3}+x^{2} y}{x-y}}: \sqrt[4]{\frac{x^{2}+x y}{x-y}} \)
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To solve these expressions, let's evaluate each step-by-step. **6. \( \sqrt[3]{\frac{x^{2}}{y}} \cdot \sqrt[12]{\frac{27 x^{2}}{16 y}}: \sqrt[6]{\frac{9 x^{5}}{4 y^{2}}} \)** First, rewrite the terms with exponents: 1. \( \sqrt[3]{\frac{x^{2}}{y}} = \left(\frac{x^{2}}{y}\right)^{\frac{1}{3}} = \frac{x^{\frac{2}{3}}}{y^{\frac{1}{3}}} \) 2. \( \sqrt[12]{\frac{27 x^{2}}{16 y}} = \left(\frac{27 x^{2}}{16 y}\right)^{\frac{1}{12}} = \frac{27^{\frac{1}{12}} x^{\frac{2}{12}}}{16^{\frac{1}{12}} y^{\frac{1}{12}}} \) Here, \( 27^{\frac{1}{12}} = 3^{\frac{3}{12}} = 3^{\frac{1}{4}} \) and \( 16^{\frac{1}{12}} = 2^{\frac{4}{12}} = 2^{\frac{1}{3}} \), so it simplifies to \( \frac{3^{\frac{1}{4}} x^{\frac{1}{6}}}{2^{\frac{1}{3}} y^{\frac{1}{12}}} \). 3. \( \sqrt[6]{\frac{9 x^{5}}{4 y^{2}}} = \left(\frac{9 x^{5}}{4 y^{2}}\right)^{\frac{1}{6}} = \frac{9^{\frac{1}{6}} x^{\frac{5}{6}}}{4^{\frac{1}{6}} y^{\frac{1}{3}}} \) Here, \( 9^{\frac{1}{6}} = 3^{\frac{2}{6}} = 3^{\frac{1}{3}} \) and \( 4^{\frac{1}{6}} = 2^{\frac{4}{6}} = 2^{\frac{2}{3}} \). Putting it all together, we have: \[ \text{Numerator: } \frac{x^{\frac{2}{3}}}{y^{\frac{1}{3}}} \cdot \frac{3^{\frac{1}{4}} x^{\frac{1}{6}}}{2^{\frac{1}{3}} y^{\frac{1}{12}}} \] The numerator becomes: \[ \frac{x^{\frac{2}{3} + \frac{1}{6}} \cdot 3^{\frac{1}{4}}}{y^{\frac{1}{3} + \frac{1}{12}} \cdot 2^{\frac{1}{3}}} \] To simplify exponents: 1. \( \frac{2}{3} = \frac{4}{6} \); thus \( \frac{2}{3} + \frac{1}{6} = \frac{5}{6} \) 2. For \( y: \frac{1}{3} + \frac{1}{12} = \frac{4}{12} + \frac{1}{12} = \frac{5}{12} \) So, the numerator is: \[ \frac{3^{\frac{1}{4}} x^{\frac{5}{6}}}{y^{\frac{5}{12}} \cdot 2^{\frac{1}{3}}} \] Now, dividing by the denominator gives: \[ \frac{\frac{3^{\frac{1}{4}} x^{\frac{5}{6}}}{y^{\frac{5}{12}} \cdot 2^{\frac{1}{3}}}}{\frac{3^{\frac{1}{3}} x^{\frac{5}{6}}}{2^{\frac{2}{3}} y^{\frac{2}{3}}}} \implies \frac{3^{\frac{1}{4} - \frac{1}{3}}}{2^{\frac{1}{3} - \frac{2}{3}}} \cdot \frac{y^{\frac{2}{3}}}{y^{\frac{5