Let's solve the given expression step by step.

The given expression is:

$\frac{\left( 2x^{-3} y^4 \right)^3 \left( \frac{x^2 y^{-3}}{4x y^5} \right)^{-2}}{16x^{-6} y^{12}}$

Step 1: Simplify $\left( 2x^{-3} y^4 \right)^3$

Apply the power rule $(a^m b^n)^p = a^{mp} b^{np}$:

$\left( 2x^{-3} y^4 \right)^3 = 2^3 x^{-9} y^{12} = 8x^{-9} y^{12}$

Step 2: Simplify $\left( \frac{x^2 y^{-3}}{4x y^5} \right)^{-2}$

First, simplify the fraction inside the parentheses:

$\frac{x^2 y^{-3}}{4x y^5} = \frac{x^2}{4x} \cdot \frac{y^{-3}}{y^5} = \frac{x^1}{4} \cdot y^{-8} = \frac{x y^{-8}}{4}$

Now, apply the exponent of $-2$ to this fraction:

$\left( \frac{x y^{-8}}{4} \right)^{-2} = \frac{4^2}{(x y^{-8})^2} = \frac{16}{x^2 y^{-16}} = \frac{16 y^{16}}{x^2}$

Step 3: Multiply the results from Step 1 and Step 2

Multiply the two simplified terms:

$8x^{-9} y^{12} \cdot \frac{16 y^{16}}{x^2} = \frac{8 \cdot 16 \cdot y^{12} \cdot y^{16}}{x^{-9} \cdot x^2} = \frac{128 y^{28}}{x^{-7}}$

Since $x^{-9} \cdot x^2 = x^{-7}$, and simplifying $x^{-7}$ moves it to the numerator as $x^7$:

$128 x^7 y^{28}$

Step 4: Divide by the denominator $16x^{-6} y^{12}$

Now divide the result from Step 3 by $16x^{-6} y^{12}$:

$\frac{128 x^7 y^{28}}{16 x^{-6} y^{12}} = \frac{128}{16} \cdot \frac{x^7}{x^{-6}} \cdot \frac{y^{28}}{y^{12}} = 8 \cdot x^{7 - (-6)} \cdot y^{28 - 12}$

This simplifies to:

$8 x^{13} y^{16}$

Final Answer:

$8 x^{13} y^{16}$

Let me know if you would like more details or have any other questions!

Here are five follow-up questions related to this topic:

1. How do you simplify expressions involving negative exponents?
2. Can you apply the power of a quotient rule to another example?
3. How does multiplying terms with exponents work?
4. What happens if a base is raised to a negative power and divided?
5. How does exponentiation affect both the numerator and denominator in a fraction?

Tip: Always simplify expressions inside parentheses before applying any exponent rules!