To simplify the given expression:

$\frac{x^2 - x - 6}{2x^2 + 7x + 6} \cdot \frac{2x^2 - x - 6}{x^2 + x - 6}$

Step 1: Factorize each quadratic expression

First fraction:

1. $x^2 - x - 6$: Factor this as $(x - 3)(x + 2)$.

2. $2x^2 + 7x + 6$: Factor this as $(2x + 3)(x + 2)$.

Second fraction:

1. $2x^2 - x - 6$: Factor this as $(2x + 3)(x - 2)$.

2. $x^2 + x - 6$: Factor this as $(x - 2)(x + 3)$.

Now, the expression becomes:

$\frac{(x - 3)(x + 2)}{(2x + 3)(x + 2)} \cdot \frac{(2x + 3)(x - 2)}{(x - 2)(x + 3)}$

Step 2: Cancel common factors

• $(x + 2)$ appears in the numerator and denominator of the first fraction, so cancel it out.
• $(2x + 3)$ appears in both the numerator and denominator of the second fraction, so cancel it out.
• $(x - 2)$ appears in the numerator and denominator of the second fraction, so cancel it out.
• $(x + 3)$ appears in the numerator and denominator of the second fraction, so cancel it out.

After cancellation, we are left with:

$\frac{x - 3}{x - 3}$

Step 3: Final simplification

If $x \neq 3$, $\frac{x - 3}{x - 3} = 1$.

Thus, the simplified expression is:

$1$

Conclusion:

The simplified expression is $\boxed{1}$, assuming $x \neq 2, -2, -3, 3$ to avoid division by zero.

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Would you like further details on any step?

Here are some related questions:

1. How do you factor quadratic expressions?
2. Why do we need to check for restrictions on the variable?
3. Can we apply this method to more complex rational expressions?
4. How would the answer change if some factors didn't cancel out?
5. How do we identify when two expressions are identical after simplification?

Tip: Always remember to check for values that make the denominator zero after simplifying!