Answer
$(9+b^{2k})(3+b^{k})(3-b^{k})$
Work Step by Step
Using $x^2-y^2=(x+y)(x-y)$ or the factoring of the difference of 2 squares, then the factored form of the given expression, $
81-b^{4k}
,$ is
\begin{array}{l}
(9+b^{2k})(9-b^{2k})
\\\\=
(9+b^{2k})(3+b^{k})(3-b^{k})
\end{array}