#### Answer

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

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^8-1
,$ use the factoring of the difference of $2$ squares repetitively until none of the resulting expressions is still a difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The expressions $
x^8
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^8-1
,$ 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^4)^2-(1)^2
\\\\=
(x^4+1)(x^4-1)
.\end{array}
The expressions $
x^4
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^4-1
,$ 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^4+1)[(x^2)^2-(1)^2]
\\\\=
(x^4+1)[(x^2+1)(x^2-1)]
\\\\=
(x^4+1)(x^2+1)(x^2-1)
.\end{array}
The expressions $
x^2
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^2-1
,$ 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^4+1)(x^2+1)[(x)^2-(1)^2]
\\\\=
(x^4+1)(x^2+1)[(x+1)(x-1)]
\\\\=
(x^4+1)(x^2+1)(x+1)(x-1)
.\end{array}