Answer
$81r^2-18rs+s^2-4$
Work Step by Step
Using $(a+b)(a-b)=a^2-b^2$ or the special product of the sum and difference of like terms, the product of the given expression, $
[(9r-s)+2][(9r-s)-2
,$ is
\begin{array}{l}\require{cancel}
(9r-s)^2-(2)^2
\\\\=
(9r-s)^2-4
.\end{array}
Using $(a\pm b)^2=a^2\pm2ab+b^2$ or the square of a binomial, the expression, $
(9r-s)^2-4
,$ is equivalent to
\begin{array}{l}\require{cancel}
(9r)^2-2(9r)(s)+(s)^2-4
\\\\=
81r^2-18rs+s^2-4
.\end{array}