Answer
$(z+1)(z-1)(z^4+z^2+1)$
Work Step by Step
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}
