Answer
$(3x-5)(9x^2-3x+7)$
Work Step by Step
With $27=3^3$, the given expression is equivalent to:
$=(3x-2)^3-3^3$
RECALL:
The difference of two cubes $a^3-b^3$ can be factored using the formula:
$a^3-b^3=(a-b)(a^2+ab+b^2)$
Factor the given difference of two cubes using the formula above with $a=3x-2$ and $b=3$ to obtain:
$(3x-2)^3-3^3
\\=[(3x-2)-3][(3x-2)^2+(3x-2)(3)+3^2]
\\=(3x-2-3)[(9x^2-2(3x)(2)+2^2)+(9x-6)+9]
\\=(3x-5)[9x^2-12x+4+9x-6+9]$
Combine like terms to obtain:
$=(3x-5)[9x^2+(-12x+9x)+(4-6+9)]
\\=(3x-5)(9x^2-3x+7)$
