#### Answer

$(a-2)(a-3)(a+3)$

#### Work Step by Step

Rearranging the terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
18+a^3-9a-2a^2
\\\\=
a^3-2a^2-9a+18
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
a^3-2a^2-9a+18
\\\\=
(a^3-2a^2)-(9a-18)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
a^2(a-2)-9(a-2)
.\end{array}
Factoring the $GCF=
(a-2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(a-2)(a^2-9)
.\end{array}
The expressions $
a^2
$ and $
9
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
a^2-9
,$ 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}
(a-2)(a^2-9)
\\\\=
(a-2)[(a)^2-(3)^2]
\\\\=
(a-2)(a-3)(a+3)
.\end{array}