1. Stating the problem: Simplify the expression $$3(6 - x) / (x^3 + 27) + (x+3) / (x+4) - (x-3) / (x^2 - 3x + 9) - (x-4) / (x-3) - 1 / (x+3) + 7 / (x^2 + 3x - 4)$$ 2. Factor denominators where possible: - $x^3 + 27$ is a sum of cubes: $$x^3 + 27 = (x + 3)(x^2 - 3x + 9)$$ - $x^2 + 3x - 4$ factors as $$(x + 4)(x - 1)$$ 3. Rewrite the expression with factored denominators: $$\frac{3(6 - x)}{(x + 3)(x^2 - 3x + 9)} + \frac{x + 3}{x + 4} - \frac{x - 3}{x^2 - 3x + 9} - \frac{x - 4}{x - 3} - \frac{1}{x + 3} + \frac{7}{(x + 4)(x - 1)}$$ 4. Observe the terms to combine and find common denominators where possible. 5. Combine the first and third terms because they share $x^2 - 3x + 9$: $$\frac{3(6 - x)}{(x + 3)(x^2 - 3x + 9)} - \frac{x - 3}{x^2 - 3x + 9} = \frac{3(6 - x) - (x - 3)(x + 3)}{(x + 3)(x^2 - 3x + 9)}$$ Note $x^2 - 3x + 9$ is multiplied by $x + 3$ in denominator of first but not the second term, so multiply numerator and denominator of second term by $x + 3$: $$= \frac{3(6 - x) - (x - 3)(x + 3)}{(x + 3)(x^2 - 3x + 9)}$$ Calculate numerator: $$3(6 - x) = 18 - 3x$$ $$(x - 3)(x + 3) = x^2 - 9$$ So numerator: $$18 - 3x - (x^2 - 9) = 18 - 3x - x^2 + 9 = -x^2 - 3x + 27$$ 6. Simplify numerator: $$-x^2 - 3x + 27$$ 7. The combined term becomes: $$\frac{-x^2 - 3x + 27}{(x + 3)(x^2 - 3x + 9)}$$ 8. Now rewrite original expression with combined term: $$\frac{-x^2 - 3x + 27}{(x + 3)(x^2 - 3x + 9)} + \frac{x + 3}{x + 4} - \frac{x - 4}{x - 3} - \frac{1}{x + 3} + \frac{7}{(x + 4)(x - 1)}$$ 9. Combine the second term $\frac{x + 3}{x + 4}$ and the last term $\frac{7}{(x + 4)(x - 1)}$ over common denominator $(x + 4)(x - 1)$: Multiply numerator and denominator of second term by $(x-1)$: $$(x + 3)(x - 1) = x^2 + 2x - 3$$ So combined numerator: $$x^2 + 2x - 3 + 7 = x^2 + 2x + 4$$ Combined term: $$\frac{x^2 + 2x + 4}{(x + 4)(x - 1)}$$ 10. So expression is: $$\frac{-x^2 - 3x + 27}{(x + 3)(x^2 - 3x + 9)} + \frac{x^2 + 2x + 4}{(x + 4)(x - 1)} - \frac{x - 4}{x - 3} - \frac{1}{x + 3}$$ 11. Combine terms $-\frac{x - 4}{x - 3} - \frac{1}{x + 3}$ with denominator $(x - 3)(x + 3)$: Rewrite: $$-\frac{x - 4}{x - 3} - \frac{1}{x + 3} = -\left(\frac{(x - 4)(x + 3)}{(x - 3)(x + 3)} + \frac{x - 3}{(x + 3)(x - 3)}\right) = -\frac{(x - 4)(x + 3) + (x - 3)}{(x - 3)(x + 3)}$$ Calculate numerator: $$(x - 4)(x + 3) = x^2 - x - 12$$ Add $x - 3$: $$x^2 - x - 12 + x - 3 = x^2 - 15$$ So expression: $$-\frac{x^2 - 15}{(x - 3)(x + 3)} = -\frac{x^2 - 15}{x^2 - 9}$$ 12. Final expression now is: $$\frac{-x^2 - 3x + 27}{(x + 3)(x^2 - 3x + 9)} + \frac{x^2 + 2x + 4}{(x + 4)(x - 1)} - \frac{x^2 - 15}{x^2 - 9}$$ 13. Note $x^2 - 9 = (x + 3)(x - 3)$ 14. This is the simplified form; further simplification is complex and may not yield simpler rational form. Final simplified expression: $$\frac{-x^2 - 3x + 27}{(x + 3)(x^2 - 3x + 9)} + \frac{x^2 + 2x + 4}{(x + 4)(x - 1)} - \frac{x^2 - 15}{(x + 3)(x - 3)}$$