Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 28

$(y-1+3x)(y^2-2y+1-3xy+3x+9x^2)$

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of two cubes, then, \begin{array}{l} (y-1)^3+27x^3 \\= [(y-1)+(3x)][(y-1)^2-(y-1)(3x)+(3x)^2] \\= [y-1+3x][(y^2-2y+1)-3xy+3x+9x^2] \\= (y-1+3x)(y^2-2y+1-3xy+3x+9x^2) .\end{array}