To factor the given expression:

$\frac{(1 + x^2)^8 \cdot 4 \cdot (2 - x^4)^3 \cdot (-4x^3) - (2 - x^4)^4 \cdot 8 \cdot (1 + x^2)^7 \cdot 2x}{((1 + x^2)^8)^2}$

let's break it down step by step:

Step 1: Simplify the denominator

$((1 + x^2)^8)^2 = (1 + x^2)^{16}$ So the denominator simplifies to: $(1 + x^2)^{16}$

Step 2: Expand and factor terms in the numerator

We'll look at the two terms separately.

First term in the numerator:

$(1 + x^2)^8 \cdot 4 \cdot (2 - x^4)^3 \cdot (-4x^3)$ This simplifies to: $-16x^3(1 + x^2)^8(2 - x^4)^3$

Second term in the numerator:

$(2 - x^4)^4 \cdot 8 \cdot (1 + x^2)^7 \cdot 2x$ This simplifies to: $16x(1 + x^2)^7(2 - x^4)^4$

Step 3: Combine both terms

Now, the numerator becomes: $-16x^3(1 + x^2)^8(2 - x^4)^3 - 16x(1 + x^2)^7(2 - x^4)^4$ We can factor out the common terms: $-16x(1 + x^2)^7(2 - x^4)^3$ The remaining part is: $x^2(1 + x^2) - (2 - x^4)$ Simplifying this part: $x^2(1 + x^2) - (2 - x^4) = x^2 + x^4 - 2 + x^4 = 2x^4 + x^2 - 2$

Step 4: Final expression

The factored numerator is: $-16x(1 + x^2)^7(2 - x^4)^3(2x^4 + x^2 - 2)$

Step 5: Divide by the denominator

Now divide by the denominator $(1 + x^2)^{16}$: $\frac{-16x(1 + x^2)^7(2 - x^4)^3(2x^4 + x^2 - 2)}{(1 + x^2)^{16}} = \frac{-16x(2 - x^4)^3(2x^4 + x^2 - 2)}{(1 + x^2)^9}$

Final Answer:

$\boxed{\frac{-16x(2 - x^4)^3(2x^4 + x^2 - 2)}{(1 + x^2)^9}}$

Would you like a more detailed explanation for any specific step, or do you have further questions?

Here are 5 related questions:

1. How do you factor a polynomial with multiple variables?
2. What are the common techniques for simplifying rational expressions?
3. How does factoring help solve algebraic equations?
4. Can you explain how to deal with higher-degree polynomials like $x^4$?
5. Why is factoring important in calculus, particularly with derivatives?

Tip: Always check for common terms in the numerator before expanding to simplify complex expressions.