This is the given problem:

$\left( x^{-3} x^2 \right)^{-1} \cdot \left( x^{-2} x^3 \right)^{-1}.$

Step-by-Step Solution:

1. Simplify the terms inside the parentheses: Using the property of exponents $x^a \cdot x^b = x^{a+b}$:

   • For the first term: $x^{-3} x^2 = x^{-3+2} = x^{-1}.$ Therefore, the first term becomes: $\left( x^{-3} x^2 \right)^{-1} = \left( x^{-1} \right)^{-1}.$

   • For the second term: $x^{-2} x^3 = x^{-2+3} = x^1 = x.$ Therefore, the second term becomes: $\left( x^{-2} x^3 \right)^{-1} = \left( x \right)^{-1}.$

2. Apply the property $(x^a)^b = x^{a \cdot b}$:

   • For the first term: $\left( x^{-1} \right)^{-1} = x^{-1 \cdot -1} = x^1 = x.$

   • For the second term: $\left( x \right)^{-1} = x^{-1}.$

3. Combine the simplified terms: Now the expression becomes: $x \cdot x^{-1}.$

4. Simplify the product: Again using the property $x^a \cdot x^b = x^{a+b}$: $x \cdot x^{-1} = x^{1 + (-1)} = x^0.$

5. Simplify $x^0$: By definition, $x^0 = 1$ (for $x \neq 0$).

Final Answer:

$\boxed{1}.$

Would you like a more detailed breakdown or have any questions about the steps? Here are five related questions to expand your understanding:

1. What are the laws of exponents, and how do they apply to combining terms like $x^a \cdot x^b$?
2. How do negative exponents relate to reciprocal terms, such as $x^{-1} = \frac{1}{x}$?
3. Why does $x^0 = 1$ for any $x \neq 0$?
4. How would the solution change if the exponents were fractions instead of integers?
5. What is the difference between simplifying and solving in algebraic contexts?

Tip: Always simplify inside parentheses first before dealing with exponents applied to the entire term!