Answer
$ x(x^{2}+1)(x-1)(x+1)$
Work Step by Step
$ x^{5}-x=\quad $
Factor out $ x $
$=x(x^{4}-1)$
The parentheses hold a difference of squares, $(x^{2})^{2}-1^{2}$
$=x(x^{2}+1)(x^{2}-1)$
The first parentheses are a sum of squares, which is prime; the second parentheses hold another difference of squares $ x^{2}-1^{2}$,
$=x(x^{2}+1)(x-1)(x+1)$
