Answer
$(3r^2+1)(9r^4-3r^2+1)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
27r^6+1
,$ use the factoring of the sum/difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
Using the factoring of the sum/difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(3r^2+1)[(3r^2)^2-(3r^2)(1)+(1)^2]
\\\\=
(3r^2+1)(9r^4-3r^2+1)
.\end{array}