Answer
$\frac{1}{(x^{2}-3x+9)} ; x \ne -3$
Work Step by Step
$\frac{(x^{2}+6x+9)}{(x^{3}+27)}. \frac{1}{(x+3)}$
$(x^{2}+6x+9)$ is in the form of $(a+b)^{2}$
$[(a+b)^{2} = a^{2} + 2ab +b^{2}]$
$[(x+3)^{2} = x^{2} + 2x(3) +3^{2} = x^{2}+6x+9]$
$x^{3}+27$ can be written as $x^{3}+3^{3}$
Factors of $x^{3}+3^{3}$ are $(x+3)(x^{2}-3x+9)$
Because $A^{3}+B^{3}$ = $(A+B)(A^{2}-AB+B^{2})$
So,
$\frac{(x^{2}+6x+9)}{(x^{3}+27)}. \frac{1}{(x+3)} $ = $\frac{(x+3)(x+3)}{(x+3)(x^{2}-3x+9)}. \frac{1}{(x+3)}$
If $x=-3$, it makes the denominator 0, so we exclude -3 for $x$.
$=\frac{(x+3)(x+3)}{(x+3)(x^{2}-3x+9)}. \frac{1}{(x+3)} ; x\ne-3$
Divide numerator and denominator by common factors.
$=\frac{1}{(x^{2}-3x+9)} ; x\ne-3;$
