Appendix A - Review - A.4 Factoring Polynomials - A.4 Assess Your Understanding - Page A40: 38

$ ( 4-3x)(9x^2 +12x +16)$

Since $64-27x^3$ is a difference of the two cubes $4^3$ and $(3x)^3$, then by making use of the difference of two cubes $$a^3 -b^3 = (a - b)(a^2 +ab + b^2),$$ we have \begin{align*} 64-27x^3&=4^3-(3x)^3\\ &= (4-3x)(4^2 +4(3x) + (3x)^2)\\ &= ( 4-3x)(16 +12x +9x^2)\\ &= ( 4-3x)(9x^2 +12x +16) \end{align*}