Question

(\frac{x + 2}{x - 3},\frac{x^2 - 9}{x^2 - 3x -}) (\frac{x^3 - 27}{x^2 - 9}cdot\frac{2x + 6}{x + 6}=)

Explanation:

Step1: Factor all polynomial terms

• $x^3 - 27 = (x-3)(x^2+3x+9)$ (difference of cubes)
• $x^2 - 9 = (x-3)(x+3)$ (difference of squares)
• $2x + 6 = 2(x+3)$ (factor out GCF)

Step2: Rewrite product with factored forms

$\frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)} \cdot \frac{2(x+3)}{x+6}$

Step3: Cancel common factors

Cancel $(x-3)$ and $(x+3)$ from numerator and denominator:
$\frac{x^2+3x+9}{1} \cdot \frac{2}{x+6}$

Step4: Multiply remaining terms

$\frac{2(x^2+3x+9)}{x+6}$

Answer:

$\frac{2x^2+6x+18}{x+6}$ (or $\frac{2(x^2+3x+9)}{x+6}$)

Note: The first expression in the image is incomplete, so only the fully visible lower rational expression was simplified.