Let's solve the expression:

$\frac{\left(\frac{a^2 - 1}{a + 1}\right)}{\left(\frac{a}{a^2 - 1}\right)} \cdot \frac{1}{\left(a - \frac{1}{a}\right)}$

Step 1: Simplify the first fraction

The expression inside the first fraction can be simplified. We know that:

$a^2 - 1 = (a - 1)(a + 1)$

So, the first fraction becomes:

$\frac{(a - 1)(a + 1)}{a + 1}$

Canceling the common term $(a + 1)$:

$= a - 1$

Now, we have:

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

Step 2: Simplify the second fraction

We can rewrite the denominator:

$\frac{a}{a^2 - 1} = \frac{a}{(a - 1)(a + 1)}$

Now the expression becomes:

$\frac{a - 1}{\frac{a}{(a - 1)(a + 1)}} \cdot \frac{1}{a - \frac{1}{a}}$

Step 3: Simplify the main fraction

We can simplify the first part by multiplying by the reciprocal of the denominator:

$\left(a - 1\right) \cdot \frac{(a - 1)(a + 1)}{a} = \frac{(a - 1)^2(a + 1)}{a}$

So now the expression becomes:

$\frac{(a - 1)^2(a + 1)}{a} \cdot \frac{1}{a - \frac{1}{a}}$

Step 4: Simplify the second term

We need to simplify $a - \frac{1}{a}$. This expression can be rewritten as:

$a - \frac{1}{a} = \frac{a^2 - 1}{a}$

Step 5: Substitute and simplify

Now substitute back into the main expression:

$\frac{(a - 1)^2(a + 1)}{a} \cdot \frac{a}{a^2 - 1}$

Since $a^2 - 1 = (a - 1)(a + 1)$, we can simplify the product:

$= \frac{(a - 1)^2(a + 1)}{a} \cdot \frac{a}{(a - 1)(a + 1)}$

Cancel out $(a - 1)(a + 1)$ and the $a$'s, and we are left with:

$a - 1$

Final Answer:

$a - 1$

Would you like more details or have any questions?

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Here are some related questions to expand on this topic:

1. How can you simplify rational expressions with higher powers of $a$?
2. What happens if you change $a^2 - 1$ to a different polynomial?
3. How would the steps change if we had a more complex denominator in the original problem?
4. How does factoring help in simplifying rational expressions?
5. Can you explain the concept of multiplying by the reciprocal in detail?

Tip: Always look for common factors in rational expressions—they simplify the problem significantly!