Question

simplify.\\(\frac{x - 8}{x^2 - 16x + 64}\\)

Explanation:

Step1: Factor the denominator

Notice that $x^2 - 16x + 64$ is a perfect square trinomial:
$x^2 - 16x + 64 = (x-8)^2$

Step2: Cancel common factors

The numerator is $x-8$, and the denominator is $(x-8)^2$. Cancel one $(x-8)$ term (where $x
eq 8$ to avoid division by zero):
$\frac{x-8}{(x-8)^2} = \frac{1}{x-8}$

Answer:

$\frac{1}{x-8}$ (where $x
eq 8$)