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

$(z+1)(z-1)(z^4+z^2+1)$
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $z^6-1 ,$ is \begin{array}{l} (z^2-1)[ (z^2)^2-(z^2)(-1)+(-1)^2] \\\\= (z^2-1)(z^4+z^2+1) .\end{array} Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, then the complete factored form of the expression above is \begin{array}{l} (z+1)(z-1)(z^4+z^2+1) .\end{array}