Answer
$\dfrac{x^{2}-16}{x^{2}-8x+16}=\dfrac{x+4}{x-4}$
Work Step by Step
$\dfrac{x^{2}-16}{x^{2}-8x+16}$
Factor the numerator:
$\dfrac{x^{2}-16}{x^{2}-8x+16}=\dfrac{(x-4)(x+4)}{x^{2}-8x+16}=...$
Factor the perfect square trinomial in the denominator. Then, simplify:
$...=\dfrac{(x-4)(x+4)}{(x-4)^{2}}=\dfrac{x+4}{x-4}$
