[tex]\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\\ \quad \\
\textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3 \\\\
-------------------------------\\\\[/tex]
[tex]\bf \cfrac{x-5}{x^3+27}+\cfrac{2}{x^2-9}\implies \cfrac{x-5}{x^3+3^3}+\cfrac{2}{x^2-3^2}
\\\\\\
\cfrac{x-5}{(x+3)(x^2-3x+3^2)}+\cfrac{2}{(x-3)(x+3)}\\\\\\
\cfrac{x-5}{(x+3)(x^2-3x+9)}+\cfrac{2}{(x-3)(x+3)}
\\\\\\
\textit{so, we'll use the LCD of }(x-3)(x+3)(x^2-3x+9)
\\\\\\
\cfrac{(x-3)(x-5)~~+~~(x^2-3x+9)2}{(x-3)(x+3)(x^2-3x+9)}
\\\\\\
\cfrac{x^2-8x+15~~+~~2x^2-6x+18}{(x-3)(x+3)(x^2-3x+9)}\implies \cfrac{3x^2-14x+33}{(x-3)(x+3)(x^2-3x+9)}[/tex]
