find the values of \( m \) and \( n \) such that \( \frac{a^{n}(\sqrt[4]{b})^{6 m}}{\sqrt[n]{a^{3 n}} b^{1-2 n}}=a^{2} b^{3} \)

To solve the equation \[ \frac{a^{n}(\sqrt[4]{b})^{6 m}}{\sqrt[n]{a^{3 n}} b^{1-2 n}}=a^{2} b^{3}, \] let's simplify the left-hand side step by step. First, we can express \(\sqrt[4]{b}\) and \(\sqrt[n]{a^{3n}}\) in terms of exponents: \[ \sqrt[4]{b} = b^{1/4}, \] thus \[ (\sqrt[4]{b})^{6m} = (b^{1/4})^{6m} = b^{\frac{6m}{4}} = b^{\frac{3m}{2}}. \] Now for \(\sqrt[n]{a^{3n}}\): \[ \sqrt[n]{a^{3n}} = (a^{3n})^{1/n} = a^{3}. \] Now, substitute these back into the expression: \[ \frac{a^{n} b^{\frac{3m}{2}}}{a^{3} b^{1 - 2n}}. \] This can be simplified further: \[ = \frac{a^{n}}{a^{3}} \cdot \frac{b^{\frac{3m}{2}}}{b^{1-2n}} = a^{n - 3} b^{\frac{3m}{2} - (1 - 2n)}. \] This simplifies to: \[ = a^{n - 3} b^{\frac{3m}{2} - 1 + 2n}. \] Now, we set this equal to the right-hand side, \( a^2 b^3 \): \[ a^{n - 3} b^{\frac{3m}{2} - 1 + 2n} = a^{2} b^{3}. \] This gives us two separate equations: 1. For the \(a\) terms: \[ n - 3 = 2 \Rightarrow n = 2 + 3 = 5. \] 2. For the \(b\) terms: \[ \frac{3m}{2} - 1 + 2n = 3. \] Substituting \(n = 5\) into the equation for the \(b\) terms: \[ \frac{3m}{2} - 1 + 2(5) = 3, \] which simplifies to: \[ \frac{3m}{2} - 1 + 10 = 3 \Rightarrow \frac{3m}{2} + 9 = 3 \Rightarrow \frac{3m}{2} = 3 - 9 = -6. \] Now solving for \(m\): \[ 3m = -12 \Rightarrow m = -4. \] Thus, the values of \( m \) and \( n \) are: \[ \boxed{m = -4 \text{ and } n = 5}. \]