Answer
$(x+3)(x-3)(x^2-3x+9)(x^2+3x+9)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^6-729
,$ use the factoring of the difference of $2$ squares. Then use the factoring of the sum or difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
The expressions $
x^6
$ and $
729
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^6-729
,$ 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^3)^2-(27)^2
\\\\=
(x^3+27)(x^3-27)
.\end{array}
The expressions $
x^3
$ and $
27
$ are both perfect cubes (the cube root is exact). Hence, $
x^3+27
$ is a $\text{
sum
}$ 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}
[(x)^3+(3)^3](x^3-27)
\\\\=
(x+3)[(x)^2-x(3)+(3)^2](x^3-27)
\\\\=
(x+3)(x^2-3x+9)(x^3-27)
.\end{array}
The expressions $
x^3
$ and $
27
$ are both perfect cubes (the cube root is exact). Hence, $
x^3-27
$ 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}
(x+3)(x^2-3x+9)[(x)^3-(3)^3]
\\\\=
(x+3)(x^2-3x+9)(x-3)[(x)^2+x(3)+(3)^2]
\\\\=
(x+3)(x^2-3x+9)(x-3)(x^2+3x+9)
\\\\=
(x+3)(x-3)(x^2-3x+9)(x^2+3x+9)
.\end{array}
