We are given the expression:

$(a - b) \left( a^{-1} \cdot b \cdot a b^{-1} \right)^{-1}$

To simplify this, let's break it down step by step.

Step 1: Simplify the inner expression

First, focus on simplifying the inner part of the expression, which is:

$\left( a^{-1} \cdot b \cdot a b^{-1} \right)^{-1}$

This is an inverse of the product $a^{-1} \cdot b \cdot a \cdot b^{-1}$. Using the property of the inverse of a product, which states that:

$(ABC)^{-1} = C^{-1} \cdot B^{-1} \cdot A^{-1}$

we can apply it here. So, we first find the inverse of $a^{-1} \cdot b \cdot a \cdot b^{-1}$:

$\left( a^{-1} \cdot b \cdot a \cdot b^{-1} \right)^{-1} = b \cdot a^{-1} \cdot b^{-1} \cdot a$

Step 2: Substituting back into the main expression

Now, substitute this back into the original expression:

$(a - b) \cdot \left( b \cdot a^{-1} \cdot b^{-1} \cdot a \right)$

Step 3: Final simplification

This expression cannot be simplified further without additional context or assumptions (e.g., commutative properties, values of $a$ and $b$). It is in a simplified form:

$(a - b) \cdot \left( b \cdot a^{-1} \cdot b^{-1} \cdot a \right)$

Let me know if you'd like more details or if you have any further questions! Here are some related questions to explore:

1. How do matrix inverses work in this context?
2. Can we simplify this further if $a$ and $b$ commute?
3. What does the inverse of a product tell us about the elements in the product?
4. How do group properties affect the simplification of expressions like this?
5. Can this expression be interpreted geometrically or in terms of group actions?

Tip: When dealing with inverses of products, remember the rule: $(ABC)^{-1} = C^{-1} \cdot B^{-1} \cdot A^{-1}$, which helps simplify complex expressions involving inverses.