Answer
$(3x-5)(9x^2-3x+7)$
Work Step by Step
Since the polynomial $(3x - 2)^3 - 27$ is a difference of two cubes, we have
$$
(3x - 2)^3 - 27=(3x-2 -3)((3x-2)^2+3(3x-2)+9)\\
=(3x-5)(9x^2-12x+4+9x-6+9)\
=(3x-5)(9x^2-3x+7)
.$$
where we used the formula $(x-y)^3=(x-y)(x^2+xy+y^2)$.
The polynomial $9x^2-3x+7$ can not be factored further because it is prime.
