#### Answer

$(x^2+9)(x+3)(x-3)$

#### Work Step by Step

The expressions $
x^4
$ and $
81
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^4-81
,$ 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^2)^2-(9)^2
\\\\=
(x^2+9)(x^2-9)
.\end{array}
The expressions $
x^2
$ and $
9
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^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}
(x^2+9)(x^2-9)
\\\\=
(x^2+9)[(x)^2-(3)^2]
\\\\=
(x^2+9)[(x+3)(x-3)]
\\\\=
(x^2+9)(x+3)(x-3)
.\end{array}