(x^2 - 9)/(x^2 - 4) / (x^2 -x - 6)/(x^2 + x-6)?

(x+3)^2 / (x +2)^2

(x - 3)²
----------
(x - 2)²

Presuming you mean:

[(x² - 9) / (x² - 4)] / [(x² - x - 6) / (x² + x - 6)]

Then the first thing that I would do is turn the division of fractions into the multiplication of the reciprocal:

[(x² - 9) / (x² - 4)] * [(x² + x - 6) / (x² - x - 6)]

Now factor everything that you can, then we will see what factors cancels out:

(x + 3)(x - 3) / [(x + 2)(x - 2)] * (x + 3)(x - 2) / [(x - 3)(x + 2)]

Now we'll start cancelling out common factors that exists in any numerator and any denominator. Starting with (x - 3) in the first numerator and second denominator:

(x + 3) / [(x + 2)(x - 2)] * (x + 3)(x - 2) / (x + 2)

Next, the (x - 2) in the first denominator and second numerator:

(x + 3) / (x + 2) * (x + 3) / (x + 2)

That's all we can do, so multiply and simplify to get your solution:

(x + 3)² / (x + 2)²
(x² + 6x + 9) / (x² + 4x + 4)

Let's simplify the numerator first:
Note that x^2 - 9 = (x - 3)(x + 3) using the difference of two squares
Also, x^2 - 4 = (x - 2)(x + 2) by the same principle.

So (x^2 - 9)/(x^2 - 4) = (x - 3)(x + 3) / (x - 2)(x + 2)

Now for the denominator:
Factorize x^2 - x - 6:
x^2 - x - 6
x^2 - 3x + 2x - 6
x (x - 3) + 2 (x - 3)
(x - 3) (x + 2)

Now factorize x^2 + x - 6:
x^2 + x - 6
x^2 + 3x - 2x - 6
x (x + 3) - 2 (x + 3)
(x + 3) (x - 2)

So (x^2 -x - 6)/(x^2 + x-6) = (x - 3)(x + 2) / (x + 3)(x - 2)

Now we can simplify the entire fraction:

[(x - 3)(x + 3) / (x - 2)(x + 2)] / [(x - 3)(x + 2) / (x + 3)(x - 2)]

Since the denominator is a fraction, find the reciprocal and multiply the the numerator:

[(x - 3)(x + 3) / (x - 2)(x + 2)] [(x + 3)(x - 2) / (x - 3)(x + 2)]

= (x - 3)(x + 3)(x + 3)(x - 2) / (x - 2)(x + 2)(x - 3)(x + 2)

The (x - 3) and (x - 2) cancels since they are common in both the numerator and the denominator.

Hence we are left with

(x + 3) (x + 3) / (x + 2) (x + 2)

Or (x + 3)^2 / (x + 2)^2

Simplify it? Is that what you want?

(x² - 9) / (x² - 4) / (x² - x - 6) / (x² + x - 6)
= (x² - 9) / [(x² - 4)(x² - x - 6)(x² + x - 6)]
= (x + 3)(x - 3) / [(x + 2)(x - 2)(x + 2)(x - 3)(x - 2)(x + 3)]
= 1 / [(x + 2)(x - 2)(x + 2)(x - 2)]
= 1 / (x² - 2)²
= 1 / (x^4 - 4x² - 4) ... where x ≠ -3, -2, 2, or 3

Follow-up:
I see that I am in a minority here. That is because, so far, I am the only one who has observed the order of operations and addressed the expression as you wrote it. I have inserted no brackets, because to do so would be to change the given expression. You gave us a combination of four quadratic expressions, three of which are divisors. That places the first one in the numerator and the other three in the denominator.