Answer
$ \displaystyle \frac{9a^{2}+6ab+4b^{2}}{b+2}$
Work Step by Step
Factor what we can:
$27a^{3}-8b^{3}$difference of cubes = $(3a)^{3}-(2b)^{3}$
$=(3a-2b)[(3a)^{2}+(3a)(2b)+(2b)^{2}]$
$=(3a-2b)(9a^{2}+6ab+4b^{2})$
$bc-b-3c+3$ = ... factor in pairs
$=b(c-1)-3(c-1)=(c-1)(b-3)$
$b^{2}-b-6=b^{2}-3b+2b-6$= ... factor in pairs
$=b(b-3)+2(b-3)=(b-3)(b+2)$
$3ac-2bc-3a+2b$= ... factor in pairs
$=c(3a-2b)-(3a-2b)=(3a-2b)(c-1)$
Rewrite the problem:
$ \displaystyle \frac{(3a-2b)(9a^{2}+6ab+4b^{2})}{(b-3)(b+2)}\cdot\frac{(c-1)(b-3)}{(3a-2b)(c-1)}=\qquad$ ... reduce common factors
$= \displaystyle \frac{(1)(9a^{2}+6ab+4b^{2})}{(b-3)(b+2)}\cdot\frac{(c-1)(b-3)}{(1)(c-1)} $
$= \displaystyle \frac{(1)(9a^{2}+6ab+4b^{2})}{(b-3)(b+2)}\cdot\frac{(1)(b-3)}{(1)(1)} $
$= \displaystyle \frac{(1)(9a^{2}+6ab+4b^{2})}{(1)(b+2)}\cdot\frac{(1)(1)}{(1)(1)} $
=$ \displaystyle \frac{9a^{2}+6ab+4b^{2}}{b+2}$
