19. \( \frac{9 x^{2}\left(x^{3}-1\right)-4\left(x^{3}-1\right)}{3 x^{2}-x-2} \div \frac{3 x^{2}+3 x+3}{9 x+6} \)

Let's simplify the given expression step by step. We start with: \[ \frac{9 x^{2}\left(x^{3}-1\right)-4\left(x^{3}-1\right)}{3 x^{2}-x-2} \div \frac{3 x^{2}+3 x+3}{9 x+6} \] First, we can factor \(x^3 - 1\): \[ x^3 - 1 = (x - 1)(x^2 + x + 1) \] Now, substitute in: \[ = \frac{(9x^2 - 4)(x^3 - 1)}{3 x^{2}-x-2} \div \frac{3 x^{2}+3 x+3}{9 x+6} \] Next, for the term \(3x^2 + 3x + 3\), we can factor out a 3: \[ = \frac{3(x^2 + x + 1)}{3(3x + 2)} = \frac{x^2 + x + 1}{3x + 2} \] Thus, our expression can now be written as: \[ = \frac{(9x^2 - 4)(x^3 - 1)}{3 x^{2}-x-2} \cdot \frac{9x + 6}{x^2 + x + 1} \] Next, factor \(9x^2 - 4\) as a difference of squares: \[ = \frac{(3x - 2)(3x + 2)(x^3 - 1)}{3 x^{2}-x-2} \cdot \frac{9x + 6}{x^2 + x + 1} \] Now, factor \(3x^2 - x - 2\): To factor \(3x^2 - x - 2\), we look for two numbers that multiply to \(-6\) (3 * -2) and add to \(-1\). These numbers are \(-3\) and \(2\). So we write: \[ 3x^2 - 3x + 2x - 2 = (3x + 2)(x - 1) \] Thus, \[ = \frac{(3x - 2)(3x + 2)(x^3 - 1)}{(3x + 2)(x - 1)} \cdot \frac{9x + 6}{x^2 + x + 1} \] Now cancel the \(3x + 2\): \[ = \frac{(3x - 2)(x^3 - 1)}{(x - 1)} \cdot \frac{9x + 6}{x^2 + x + 1} \] Next, simplify \(9x + 6\): \[ = 3(3x + 2) \] Now, combine everything: \[ = \frac{(3x - 2)(x^3 - 1) \cdot 3(3x + 2)}{(x - 1)(x^2 + x + 1)} \] Substituting \(x^3 - 1\) back gives: \[ = \frac{(3x - 2)(x - 1)(x^2 + x + 1) \cdot 3(3x + 2)}{(x - 1)(x^2 + x + 1)} \] Now we cancel \(x -1\) and \(x^2 + x + 1\): \[ = (3x - 2) \cdot 3(3x + 2) \] Finally, our complete simplified expression is: \[ = 3(3x - 2)(3x + 2) = 3(9x^2 - 4) = 27x^2 - 12 \] Thus, the final answer is: \[ \boxed{27x^2 - 12} \]