Answer:
[tex]\dfrac{x-4}{x^2+3x+9}[/tex]
Note:
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)\\a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]
Step-by-step explanation:
[tex]\dfrac{x^2-7x+12}{x^3-27}\\\\=\dfrac{x^2-4x-3x+12}{x^3-3^3}\\\\=\dfrac{x(x-4)-3(x-4)}{(x-3)(x^2+3x+3^2)}\\\\=\dfrac{(x-4)(x-3)}{(x-3)(x^2+3x+9)}\\\\=\dfrac{x-4}{x^2+3x+9}[/tex]
