Answer
$\displaystyle \frac{(x-4)(x^{2}+4)}{ x-1}$
Work Step by Step
Dividing with $\displaystyle \frac{P}{Q}$ = multiplying with $\displaystyle \frac{Q}{P}$
Rewrite the problem:
$ \displaystyle \frac{(x^{2}-16)}{1}\cdot\frac{x^{2}+4}{x^{2}+3x-4}=\qquad$ ... factor what we can
... $x^{2}-16$ = difference of squares = $(x+4)(x-4)$
... $ x^{2}+4\qquad$ is prime
... $x^{2}+3x-4=$... factors of $-4$ whose sum is $+3$ ... are $+4$ and $-1$
$=(x+4)(x-1)$
Rewrite the problem:
$ = \displaystyle \frac{(x+4)(x-4)}{1}\cdot\frac{x^{2}+4}{(x+4)(x-1)}=\qquad$ ... reduce common factors
$ = \displaystyle \frac{(1)(x-4)}{1}\cdot\frac{x^{2}+4}{(1)(x-1)}=$
= $\displaystyle \frac{(x-4)(x^{2}+4)}{ x-1}$
