#### Answer

$(3+a-b)(3-a+b)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
9-a^2+2ab-b^2
,$ group the last $3$ terms and factor the trinomial. Then use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the last $3$ terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
9-(a^2-2ab+b^2)
.\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}
9-(a-b)(a-b)
\\\\
9-(a-b)^2
.\end{array}
The expressions $
9
$ and $
(a-b)^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
9-(a-b)^2
,$ 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}
(3)^2-(a-b)^2
\\\\=
[3+(a-b)][3-(a-b)]
\\\\=
(3+a-b)(3-a+b)
.\end{array}