Answer
$\frac{1}{x+3}$.
Work Step by Step
The given expression is
$=\frac{x^2-9}{x^3-27}\div \frac{x^2+6x+9}{x^2+3x+9}$
Replace divide sign with multiplication sign and reciprocal the right hand side fraction as shown below.
$=\frac{x^2-9}{x^3-27}\times \frac{x^2+3x+9}{x^2+6x+9}$
Factor each denominator and numerator.
$=x^2-9$
$=x^2-3^2$
Use algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(x+3)(x-3)$
$=x^3-27$
$=x^3-3^3$
Use algebraic identity $a^3-b^3=(a-b)(a^2+ab+b^2)$.
$=(x-3)(x^2+3x+9)$
$=x^2+3x+9$
This is the prime polynomial.
$=x^2+6x+9$
Rewrite middle term.
$=x^2+3x+3x+9$
Group terms.
$=(x^2+3x)+(3x+9)$
Factor from each group.
$=x(x+3)+3(x+3)$
Factor out common term.
$=(x+3)(x+3)$
Plug all factors into the given expression.
$=\frac{(x+3)(x-3)}{(x-3)(x^2+3x+9)}\times \frac{x^2+3x+9}{(x+3)(x+3)}$
Cancel common terms.
$=\frac{1}{x+3}$.
