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