We need to multiply and simplify the expression $(\frac{3x^6}{2y^4})^5 \cdot (\frac{2^3y^3}{3^4x^6})$ completely, eliminating all negative exponents. We need to provide the numerator and denominator separately.

1. Problem Description

We need to multiply and simplify the expression $(\frac{3x^6}{2y^4})^5 \cdot (\frac{2^3y^3}{3^4x^6})$ completely, eliminating all negative exponents. We need to provide the numerator and denominator separately.

2. Solution Steps

First, let's simplify the first term by raising it to the power of 5:

$(\frac{3x^6}{2y^4})^5 = \frac{(3x^6)^5}{(2y^4)^5} = \frac{3^5 (x^6)^5}{2^5 (y^4)^5} = \frac{3^5 x^{30}}{2^5 y^{20}}$.

Now, let's multiply this by the second term:

$\frac{3^5 x^{30}}{2^5 y^{20}} \cdot \frac{2^3 y^3}{3^4 x^6} = \frac{3^5 x^{30} \cdot 2^3 y^3}{2^5 y^{20} \cdot 3^4 x^6}$.

We can now simplify by combining like terms:

$\frac{3^5}{3^4} = 3^{5-4} = 3^1 = 3$

$\frac{2^3}{2^5} = 2^{3-5} = 2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

$\frac{x^{30}}{x^6} = x^{30-6} = x^{24}$

$\frac{y^3}{y^{20}} = y^{3-20} = y^{-17} = \frac{1}{y^{17}}$

Putting it all together:

$\frac{3^5 x^{30} 2^3 y^3}{2^5 y^{20} 3^4 x^6} = 3 \cdot \frac{1}{2^2} \cdot x^{24} \cdot \frac{1}{y^{17}} = \frac{3x^{24}}{4y^{17}}$.

The numerator is $3x^{24}$.

The denominator is $4y^{17}$.

3. Final Answer

Numerator: $3x^{24}$

Denominator: $4y^{17}$