## College Algebra (11th Edition)

(3 - m - 2n)($m^{2}$ +4$n^{2}$ + 4mn+ 3m + 6n + 9)
27 - $(m + 2n)^{3}$ is a difference of cubes, so we write it as a difference of cubes and factor: $3^{3}$ - $(m + 2n)^{3}$ = (3 - (m + 2n))($3^{2}$ + 3(m + 2n) + $(m + 2n)^{2}$) = (3 - m - 2n)(9 + 3m + 6n + $m^{2}$ + 4mn + 4$n^{2}$) = (3 - m - 2n)($m^{2}$ +4$n^{2}$ + 4mn+ 3m + 6n + 9)