#### Answer

$(1+x^{8})(1+x^{4})(1+x^{2})(1+x)(1-x)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
1-x^{16}
,$ use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The expressions $
1
$ and $
x^{16}
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
1-x^{16}
,$ 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}
(1)^2-(x^{8})^2
\\\\=
(1+x^{8})(1-x^{8})
.\end{array}
The expressions $
1
$ and $
x^{8}
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
1-x^{8}
,$ 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}
(1+x^{8})[(1)^2-(x^{4})^2]
\\\\=
(1+x^{8})[(1+x^{4})(1-x^{4})]
\\\\=
(1+x^{8})(1+x^{4})(1-x^{4})
.\end{array}
The expressions $
1
$ and $
x^{4}
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
1-x^{4}
,$ 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}
(1+x^{8})(1+x^{4})[(1)^2-(x^{2})^2]
\\\\
(1+x^{8})(1+x^{4})[(1+x^{2})(1-x^{2})]
\\\\
(1+x^{8})(1+x^{4})(1+x^{2})(1-x^{2})
.\end{array}
The expressions $
1
$ and $
x^{2}
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
1-x^{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}
(1+x^{8})(1+x^{4})(1+x^{2})[(1)^2-(x)^{2}]
\\\\=
(1+x^{8})(1+x^{4})(1+x^{2})[(1+x)(1-x)]
\\\\=
(1+x^{8})(1+x^{4})(1+x^{2})(1+x)(1-x)
.\end{array}