Answer
$9(x+2)(3x^2+4)$
Work Step by Step
Using $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$ or the factoring of two cubes, the factored form of the given expression, $
64+(3x+2)^3
,$ is
\begin{array}{l}\require{cancel}
[4+(3x+2)][(4)^2-4(3x+2)+(3x+2)^2]
\\\\=
(4+3x+2)[16-12x-8+(3x)^2+2(3x)(2)+(2)^2]
\\\\=
(4+3x+2)[16-12x-8+9x^2+12x+4]
\\\\=
(3x+6)[9x^2+12]
\\\\=
3(x+2)3(3x^2+4)
\\\\=
9(x+2)(3x^2+4)
.\end{array}
