We are asked to simplify the expression $\frac{b^3 - b^2}{1 - b^2} \times (4b + 4)$.

1. Problem Description

We are asked to simplify the expression $\frac{b^3 - b^2}{1 - b^2} \times (4b + 4)$.

2. Solution Steps

First, we factor the numerator and denominator of the fraction:

$b^3 - b^2 = b^2(b - 1)$

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

Then, we factor the term $(4b + 4)$:

$4b + 4 = 4(b + 1)$

Now we substitute these factorizations into the original expression:

$\frac{b^2(b - 1)}{-(b - 1)(1 + b)} \times 4(b + 1)$

We can cancel the $(b - 1)$ terms:

$\frac{b^2}{-(1 + b)} \times 4(b + 1)$

$\frac{b^2}{-(b + 1)} \times 4(b + 1)$

Now we can cancel the $(b + 1)$ terms:

$\frac{b^2}{-1} \times 4$

$-b^2 \times 4$

$-4b^2$

3. Final Answer

$-4b^2$