#### Answer

$(x-y+2)(x-y-2)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^2-2xy+y^2-4
,$ group the first $3$ terms and factor the trinomial. Then use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the first $3$ terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^2-2xy+y^2)-4
.\end{array}
In the trinomial expression above, $b=
-2
,\text{ and } c=
1
.$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{
-1,-1
\right\}.$ Using these two numbers, the factored form of the expression above is
\begin{array}{l}\require{cancel}
(x-1y)(x-1y)-4
\\\\=
(x-y)^2-4
.\end{array}
The expressions $
(x-y)^2
$ and $
4
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
(x-y)^2-4
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(x-y)^2-(2)^2
\\\\=
[(x-y)+2][(x-y)-2]
\\\\=
(x-y+2)(x-y-2)
.\end{array}