#### Answer

$(1-3x^2)(1+3x^2+9x^4)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $ 1-27x^6 ,$ use the factoring of the sum or difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
The expressions $
1
$ and $
27x^6
$ are both perfect cubes (the cube root is exact). Hence, $
1-27x^6
$ is a $\text{
difference
}$ of $2$ cubes. Using the factoring of the sum or 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}
1^3-(3x^2)^3
\\\\=
(1-3x^2)(1+3x^2+9x^4)
.\end{array}