1. **State the problem:** Simplify the expression $$\left(\frac{1}{x-1} + \frac{3}{1-x^2}\right) : \left(1 - \frac{1}{x-1} + \frac{2}{x+2}\right) \cdot \frac{x^2 + 3x + 6}{x^3}.$$\n\n2. **Rewrite and factor where possible:** Note that $$1 - x^2 = (1 - x)(1 + x) = -(x-1)(x+1)$$ because $$1-x^2 = -(x^2-1) = -(x-1)(x+1).$$\n\n3. **Simplify the first big parenthesis:**\n$$\frac{1}{x-1} + \frac{3}{1-x^2} = \frac{1}{x-1} + \frac{3}{-(x-1)(x+1)} = \frac{1}{x-1} - \frac{3}{(x-1)(x+1)}.$$\nFind common denominator $(x-1)(x+1)$:\n$$\frac{1}{x-1} = \frac{x+1}{(x-1)(x+1)}.$$\nSo,\n$$\frac{x+1}{(x-1)(x+1)} - \frac{3}{(x-1)(x+1)} = \frac{x+1-3}{(x-1)(x+1)} = \frac{x-2}{(x-1)(x+1)}.$$\n\n4. **Simplify the second big parenthesis:**\n$$1 - \frac{1}{x-1} + \frac{2}{x+2}.$$\nRewrite 1 as $$\frac{(x-1)(x+2)}{(x-1)(x+2)}$$ to get common denominator $(x-1)(x+2)$:\n$$\frac{(x-1)(x+2)}{(x-1)(x+2)} - \frac{1}{x-1} + \frac{2}{x+2}.$$\nRewrite the other fractions with common denominator:\n$$- \frac{1}{x-1} = - \frac{x+2}{(x-1)(x+2)}, \quad \frac{2}{x+2} = \frac{2(x-1)}{(x-1)(x+2)}.$$\nSum all:\n$$\frac{(x-1)(x+2) - (x+2) + 2(x-1)}{(x-1)(x+2)}.$$\nExpand numerator:\n$$(x-1)(x+2) = x^2 + 2x - x - 2 = x^2 + x - 2,$$\nso numerator is\n$$x^2 + x - 2 - (x+2) + 2x - 2 = x^2 + x - 2 - x - 2 + 2x - 2 = x^2 + 2x - 6.$$\n\n5. **Rewrite the entire expression:**\n$$\frac{x-2}{(x-1)(x+1)} : \frac{x^2 + 2x - 6}{(x-1)(x+2)} \cdot \frac{x^2 + 3x + 6}{x^3}.$$\nDivision by a fraction is multiplication by its reciprocal, so\n$$= \frac{x-2}{(x-1)(x+1)} \cdot \frac{(x-1)(x+2)}{x^2 + 2x - 6} \cdot \frac{x^2 + 3x + 6}{x^3}.$$\n\n6. **Factor where possible:**\n$$x^2 + 2x - 6$$ does not factor nicely (discriminant $= 4 + 24 = 28$ not a perfect square).\n$$x^2 + 3x + 6$$ also does not factor nicely (discriminant $= 9 - 24 = -15 < 0$).\n\n7. **Cancel common factors:**\nCancel $(x-1)$ from numerator and denominator:\n$$\frac{x-2}{\cancel{(x-1)}(x+1)} \cdot \frac{\cancel{(x-1)}(x+2)}{x^2 + 2x - 6} \cdot \frac{x^2 + 3x + 6}{x^3} = \frac{(x-2)(x+2)(x^2 + 3x + 6)}{(x+1)(x^2 + 2x - 6) x^3}.$$\nNote that $(x-2)(x+2) = x^2 - 4.$\n\n8. **Final simplified expression:**\n$$\boxed{\frac{(x^2 - 4)(x^2 + 3x + 6)}{(x+1)(x^2 + 2x - 6) x^3}}.$$