Algebra: A Combined Approach (4th Edition)

$\dfrac{x-3}{x^{2}-6x+9}=\dfrac{1}{x-3}$
$\dfrac{x-3}{x^{2}-6x+9}$ The denominator is a perfect square trinomial. Factor it and then simplify: $\dfrac{x-3}{x^{2}-6x+9}=\dfrac{x-3}{(x-3)^{2}}=\dfrac{1}{x-3}$