Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$3z(z-1)(z^2+z+1)$
Factoring the $GCF= 3z^2 ,$ the given expression, $3z^5-3z^2 ,$ is equivalent to \begin{array}{l} 3z^2(z^3-1) .\end{array} Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the expression, $3z^2(z^3-1) ,$ is \begin{array}{l} 3z(z-1)[ (z)^2-(z)(-1)+(-1)^2] \\\\= 3z(z-1)(z^2+z+1) .\end{array}