Answer
$= \displaystyle \frac{1}{x^{2}-3x+9},\qquad x\neq-3$
Work Step by Step
Factor what you can:
$x^{2}+6x+9=(x+3)^{2}\qquad $... square of a sum
$ x^{3}+27= \quad$...sum of cubes = $x^{3}+3^{3},$
$=(x+3)(x^{2}-3x+9)$
... the second numerator and denominator are factored.
Expression $=\displaystyle \frac{(x+3)(x+3)}{(x+3)(x^{2}-3x+9)}\cdot\frac{1}{(x+3)}\qquad$
...exclude the values that yield 0 in the denominator:
$x\neq-3$
... cancel common factors: $x$, and $(x-3)$
$= \displaystyle \frac{1}{x^{2}-3x+9},\qquad x\neq-3$
