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

$(m+1)(m-1)(m^2-m+1)(m^2+m+1)$
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}