Answer
$\dfrac{1}{6}$
Work Step by Step
Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, then the given expression, $
\dfrac{x^2-6x-16}{2x^2-128}\cdot\dfrac{x^2+16x+64}{3x^2+30x+48}
$, simplifies to
\begin{array}{l}\require{cancel}
\dfrac{(x-8)(x+2)}{2(x^2-64)}\cdot\dfrac{(x+8)(x+8)}{3(x^2+10x+16)}
\\\\=
\dfrac{(\cancel{x-8})(\cancel{x+2})}{2(\cancel{x+8})(\cancel{x-8})}\cdot\dfrac{(\cancel{x+8})(\cancel{x+8})}{3(\cancel{x+8})(\cancel{x+2})}
\\\\=
\dfrac{1}{6}
.\end{array}
