#### Answer

$5(w^{8}+1)(w^{4}+1)(w^{2}+1)(w+1)(w-1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
5w^{16}-5
,$ factor first the $GCF.$ Then use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=5,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
5(w^{16}-1)
.\end{array}
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}
5[(w^{8})^2-(1)^2]
\\\\=
5(w^{8}+1)(w^{8}-1)
\\\\=
5(w^{8}+1)[(w^{4})^2-(1)^2]
\\\\=
5(w^{8}+1)(w^{4}+1)(w^{4}-1)
\\\\=
5(w^{8}+1)(w^{4}+1)[(w^{2})^2-(1)^2]
\\\\=
5(w^{8}+1)(w^{4}+1)(w^{2}+1)(w^{2}-1)
\\\\=
5(w^{8}+1)(w^{4}+1)(w^{2}+1)[(w)^{2}-(1)^2]
\\\\=
5(w^{8}+1)(w^{4}+1)(w^{2}+1)(w+1)(w-1)
.\end{array}