## Intermediate Algebra (6th Edition)

$x(x-1)(x^2+x+1)$
Factoring the $GCF=x$, then the given expression, $x^4-x$, is equivalent to \begin{array}{l} x(x^3-1) .\end{array} Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ or the factoring of the difference of 2 cubes, then, \begin{array}{l}\require{cancel} x(x^3-1) \\\\= x(x-1)(x^2+x+1) .\end{array}