[m^3 n^{-1/4}]^2 ÷ √[5]{32 m^4 n^{-8}}

Steps to Solve

1. Simplify the Numerator
   We start by simplifying the numerator, which is ( [m^3 n^{-1/4}]^2 ).
   Using the power of a power property ( (a^m)^n = a^{mn} ), we have: $$ [m^3 n^{-1/4}]^2 = m^{3 \cdot 2} n^{-1/4 \cdot 2} = m^6 n^{-1/2} $$

2. Simplify the Denominator
   Next, we simplify the denominator, which is ( \sqrt[5]{32 m^4 n^{-8}} ).
   This can be written as: $$ \sqrt[5]{32} = 32^{1/5} = 2^{5 \cdot 1/5} = 2 $$
   So, we have:
   $$ \sqrt[5]{32 m^4 n^{-8}} = \sqrt[5]{32} \cdot \sqrt[5]{m^4} \cdot \sqrt[5]{n^{-8}} = 2 m^{4/5} n^{-8/5} $$

3. Set up the Final Expression
   Now, substituting the simplified numerator and denominator into the full expression: $$ \frac{m^6 n^{-1/2}}{2 m^{4/5} n^{-8/5}} $$

4. Simplify the Fraction We can now simplify the fraction: $$ = \frac{m^6}{2 m^{4/5}} \cdot \frac{n^{-1/2}}{n^{-8/5}} $$
   For the base ( m ): $$ \frac{m^6}{m^{4/5}} = m^{6 - 4/5} = m^{\frac{30}{5} - \frac{4}{5}} = m^{\frac{26}{5}} $$
   For the base ( n ): $$ \frac{n^{-1/2}}{n^{-8/5}} = n^{-1/2 + 8/5} = n^{-\frac{5}{10} + \frac{16}{10}} = n^{\frac{11}{10}} $$
   So we have: $$ \frac{m^{26/5} n^{11/10}}{2} $$

5. Final Expression
   Thus, the simplified expression is: $$ \frac{m^{26/5} n^{11/10}}{2} $$