#### Answer

$(r+2)(r-2)(r^2-2r+4)(r^2+2r+4)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
r^6-64
,$ 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 $
r^6
$ and $
64
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
r^6-64
,$ 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}
(r^3)^2-(8)^2
\\\\=
(r^3+8)(r^3-8)
.\end{array}
The expressions $
r^3
$ and $
8
$ are both perfect cubes (the cube root is exact). Hence, $
r^3+8
$ 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}
[(r)^3+(2)^3](r^3-8)
\\\\=
[(r+2)(r^2-r(2)+(2)^2](r^3-8)
\\\\=
(r+2)(r^2-2r+4)(r^3-8)
.\end{array}
The expressions $
r^3
$ and $
8
$ are both perfect cubes (the cube root is exact). Hence, $
r^3-8
$ 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}
(r+2)(r^2-2r+4)[(r)^3-(2)^3]
\\\\=
(r+2)(r^2-2r+4)(r-2)[(r)^2+r(2)+(2)^2]
\\\\=
(r+2)(r^2-2r+4)(r-2)(r^2+2r+4)
\\\\=
(r+2)(r-2)(r^2-2r+4)(r^2+2r+4)
.\end{array}