#### Answer

$(a+3b)(a-3b)(a^4+9a^2b^2+81b^4)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
a^6-729b^6
,$ use the factoring of the sum or difference of $2$ cubes. Then use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The expressions $
a^6
$ and $
729b^6
$ are both perfect cubes (the cube root is exact). Hence, $
a^6-729b^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}
(a^2)^3-(9b^2)^3
\\\\=
(a^2-9b^2)[(a^2)^2+a^2(9b^2)+(9b^2)^2]
\\\\=
(a^2-9b^2)(a^4+9a^2b^2+81b^4)
.\end{array}
The expressions $
a^2
$ and $
9b^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
a^2-9b^2
,$ 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}
[(a)^2-(3b)^2](a^4+9a^2b^2+81b^4)
\\\\=
(a+3b)(a-3b)(a^4+9a^2b^2+81b^4)
.\end{array}