Simplify \( \frac{x^{3}-8}{4-x^{2}} \times \frac{x+2}{3 x^{2}+6 x+12} \)

To simplify the expression \( \frac{x^{3}-8}{4-x^{2}} \times \frac{x+2}{3 x^{2}+6 x+12} \), we can start by factoring each component. 1. Factor \( x^3 - 8 \): This is a difference of cubes, which can be factored as: \[ x^3 - 8 = (x - 2)(x^2 + 2x + 4). \] 2. Factor \( 4 - x^2 \): This is a difference of squares, and can be written as: \[ 4 - x^2 = (2 - x)(2 + x). \] However, we can rewrite \( (2 - x) \) as \( -(x - 2) \): \[ 4 - x^2 = -(x - 2)(2 + x). \] 3. Factor \( 3x^2 + 6x + 12 \): This expression can be factored by taking \( 3 \) out: \[ 3x^2 + 6x + 12 = 3(x^2 + 2x + 4). \] The quadratic \( x^2 + 2x + 4 \) does not factor further over the reals. Now we can substitute all the factors in: \[ \frac{(x - 2)(x^2 + 2x + 4)}{-(x - 2)(2 + x)} \times \frac{x + 2}{3(x^2 + 2x + 4)}. \] Next, we can cancel out common terms in the numerator and denominator: - \( (x - 2) \) cancels from the numerator and denominator. - \( (x^2 + 2x + 4) \) also cancels from the numerator and denominator. After canceling, the expression simplifies to: \[ \frac{x + 2}{-3(2 + x)}. \] Since \( 2 + x \) is the same as \( x + 2 \), we have: \[ \frac{x + 2}{-3(x + 2)}. \] Now, assuming \( x + 2 \neq 0 \), we can cancel \( (x + 2) \): \[ \frac{1}{-3} = -\frac{1}{3}. \] Thus, the simplified form of the given expression is: \[ \boxed{-\frac{1}{3}}. \]