Answer
$\dfrac{x+7}{5(x+3)(x+3)}$
Work Step by Step
Factoring the expressions and then cancelling common factors between the numerator and the denominator, then the given expression, $
\dfrac{x^2-3x+9}{5x^2-20x-105}\cdot\dfrac{x^2-49}{x^3+27}
$, simplifies to
\begin{array}{l}\require{cancel}
\dfrac{x^2-3x+9}{5(x^2-4x-21)}\cdot\dfrac{(x+7)(x-7)}{(x+3)(x^2-3x+9)}
\\\\=
\dfrac{\cancel{x^2-3x+9}}{5(\cancel{x-7})(x+3)}\cdot\dfrac{(x+7)(\cancel{x-7})}{(x+3)(\cancel{x^2-3x+9})}
\\\\=
\dfrac{x+7}{5(x+3)(x+3)}
.\end{array}
