Answer
$(p-q+5)(p^2-2pq-q^2-5p-5q+25)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
(p-q)^3+125
,$ use the factoring of the sum/difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
Using the factoring of the sum/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}
[(p-q)+5][(p-q)^2-(p-q)(5)+(5)^2]
\\\\=
(p-q+5)[(p-q)^2-5p-5q+25]
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
(p-q+5)[(p^2-2pq-q^2)-5p-5q+25]
\\\\=
(p-q+5)(p^2-2pq-q^2-5p-5q+25)
.\end{array}