b. \( \frac{6 x}{x-9} \cdot \frac{x^{2}-10 x+9}{2 x+12} \cdot \frac{x+6}{x^{4}+x^{3}-2 x^{2}} \)

To tackle the expression, let's first simplify each fraction. 1. The first fraction is \( \frac{6x}{x-9} \). 2. The second fraction requires factoring \( x^2 - 10x + 9 \), which gives \( (x-1)(x-9) \). Therefore, \( \frac{x^2-10x+9}{2x+12} = \frac{(x-1)(x-9)}{2(x+6)} \). 3. The third fraction \( \frac{x+6}{x^4+x^3-2x^2} \) can be factored. Begin by factoring out \( x^2 \) from \( x^4 + x^3 - 2x^2 = x^2(x^2 + x - 2) \). The quadratic \( x^2 + x - 2 \) factors to \( (x-1)(x+2) \). So, it becomes \( x^2 \cdot (x-1)(x+2) \), leading to \( \frac{x+6}{x^2(x-1)(x+2)} \). Now, the entire expression rewrites as: \[ \frac{6x}{x-9} \cdot \frac{(x-1)(x-9)}{2(x+6)} \cdot \frac{x+6}{x^2(x-1)(x+2)} \] Cancelling common terms gives: - Cancel \( (x-9) \) and \( (x-9) \) from the first and second fractions. - Cancel \( (x+6) \) from the second and third fractions. - Cancel \( (x-1) \) from the second and third fractions. After these cancellations, we simplify the expression to: \[ \frac{6x}{2x^2} = \frac{3}{x} \] Provided that \( x \neq 9, -6, 1, 0, -2 \) (from limits affected by the initial fractions). So, the final answer is: \[ \frac{3}{x} \] (where \( x \neq 9, -6, 1, 0, -2 \)).