To simplify the expression:

$\left(a^2 \cdot b^{-2}\right)^3 \cdot \left((2a)^3 \cdot b\right)^{-2}$

Step 1: Simplify each part individually.

For $\left(a^2 \cdot b^{-2}\right)^3$:

Apply the exponent to both $a^2$ and $b^{-2}$:

$\left(a^2 \cdot b^{-2}\right)^3 = a^{2 \cdot 3} \cdot b^{-2 \cdot 3} = a^6 \cdot b^{-6}$

For $\left((2a)^3 \cdot b\right)^{-2}$:

First, simplify $(2a)^3$ and then apply the $-2$ exponent:

$(2a)^3 = 2^3 \cdot a^3 = 8a^3$

Now apply the $-2$ exponent:

$\left(8a^3 \cdot b\right)^{-2} = (8a^3)^{-2} \cdot b^{-2} = 8^{-2} \cdot a^{-6} \cdot b^{-2} = \frac{1}{64} \cdot a^{-6} \cdot b^{-2}$

Step 2: Combine the results.

Now multiply the two parts:

$\left(a^6 \cdot b^{-6}\right) \cdot \left(\frac{1}{64} \cdot a^{-6} \cdot b^{-2}\right)$

$= \frac{1}{64} \cdot a^{6 + (-6)} \cdot b^{-6 + (-2)} = \frac{1}{64} \cdot a^0 \cdot b^{-8}$

Since $a^0 = 1$, the final simplified expression is:

$\frac{1}{64} \cdot b^{-8} = \frac{1}{64b^8}$

Final Answer:

$\boxed{\frac{1}{64b^8}}$

Would you like more details or have any questions?

Related questions:

1. How do we apply exponents when there are negative powers?
2. What are the rules for multiplying and dividing terms with exponents?
3. How can we simplify expressions involving multiple variables?
4. What happens when we raise a product to a negative exponent?
5. How do fractional exponents relate to roots?

Tip:

When simplifying, always handle powers and parentheses step-by-step, especially when dealing with negative exponents.