1. Simplify the expression \(\frac{12a^0 b^3 c^{-1}}{15 a^{-2} b^5 c^{-3}}\). 2. Simplify the expression \(\frac{(3xy^{-2})^{-2}}{3x^{-2} y}\). ### Step 1: Simplify \(\frac{12a^0 b^3 c^{-1}}{15 a^{-2} b^5 c^{-3}}\) - Recall that \(a^0 = 1\) for any \(a \neq 0\). - Use the rule \(\frac{x^m}{x^n} = x^{m-n}\). \[ \frac{12a^0 b^3 c^{-1}}{15 a^{-2} b^5 c^{-3}} = \frac{12 \cdot 1 \cdot b^3 \cdot c^{-1}}{15 \cdot a^{-2} \cdot b^5 \cdot c^{-3}} = \frac{12 b^3 c^{-1}}{15 a^{-2} b^5 c^{-3}}. \] Apply the exponent subtraction rule for each variable: \[ = \frac{12}{15} \cdot a^{0 - (-2)} \cdot b^{3 - 5} \cdot c^{-1 - (-3)} = \frac{12}{15} \cdot a^{2} \cdot b^{-2} \cdot c^{2}. \] Simplify the fraction \(\frac{12}{15}\): \[ \frac{12}{15} = \frac{\cancel{3} \times 4}{\cancel{3} \times 5} = \frac{4}{5}. \] So the simplified expression is: \[ \frac{4}{5} a^{2} b^{-2} c^{2}. \] Rewrite negative exponents as positive by moving terms: \[ \frac{4}{5} a^{2} \frac{c^{2}}{b^{2}} = \frac{4 a^{2} c^{2}}{5 b^{2}}. \] ### Step 2: Simplify \(\frac{(3xy^{-2})^{-2}}{3x^{-2} y}\) First, simplify the numerator: \[ (3xy^{-2})^{-2} = 3^{-2} x^{-2} (y^{-2})^{-2} = 3^{-2} x^{-2} y^{4}. \] Recall \(3^{-2} = \frac{1}{3^{2}} = \frac{1}{9}\). So numerator is: \[ \frac{1}{9} x^{-2} y^{4}. \] Now write the entire expression: \[ \frac{\frac{1}{9} x^{-2} y^{4}}{3 x^{-2} y} = \frac{1}{9} x^{-2} y^{4} \times \frac{1}{3 x^{-2} y} = \frac{1}{9} x^{-2} y^{4} \times \frac{1}{3} x^{2} y^{-1}. \] Multiply coefficients and variables: \[ = \frac{1}{9} \times \frac{1}{3} \times x^{-2 + 2} \times y^{4 - 1} = \frac{1}{27} x^{0} y^{3}. \] Since \(x^{0} = 1\), the expression simplifies to: \[ \frac{y^{3}}{27}. \] ### Final answers: 1. \(\frac{4 a^{2} c^{2}}{5 b^{2}}\) 2. \(\frac{y^{3}}{27}\)