Answer
$(3x-5)(9x^2-3x+7)$
Work Step by Step
Write $27$ as $3^3$ to obtain:
$(3x-2)^3-27=(3x-2)^3-3^3$
Use the special formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ where $a=(3x-2)$ and $b=3$.
$=[(3x-2)-3][(3x-2)^2+(3x-2)(3)+3^2]$
Simplify.
$=(3x-5)[(3x-2)^2+9x-6+9]$
$=(3x-5)[(3x-2)^2+9x+3]$
Use special formula $(a-b)^2=a^2-2ab+b^2$ where $a=3x$ and $b=2$.
$=(3x-5)[(3x)^2-2(3x)(2)+(2)^2+9x+3]$
Simplify.
$=(3x-5)[9x^2-12x+4+9x+3]$
$=(3x-5)(9x^2-3x+7)$
Hence, the completely factored form of the given expression is $(3x-5)(9x^2-3x+7)$.
