Chapter 6 - Factoring, Solving Equations, and Problem Solving - 6.2 - Factoring the Difference of Two Squares - Problem Set 6.2 - Page 249: 95

$(3x-2y)(9x^2+6xy+4y^2)$

Using $a^3-b^3=(a-b)(a^2+ab+b^2)$, or the factoring of the difference of $2$ cubes, the factored form of the given expression, $ 27x^3-8y^3 ,$ is \begin{array}{l}\require{cancel} (3x-2y)[(3x)^2+3x(2y)+(2y)^2] \\\\= (3x-2y)(9x^2+6xy+4y^2) .\end{array}