Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 310: 66

$(r+s+3)(r^2+2rs+s^2-3r-3s+9)$

Using $a^3+b^3=(a+b)(a^2-2ab+b^2)$ or the factoring of the sum of 2 cubes, then, \begin{array}{l} (r+s)^3+27 \\= [(r+s)+(3)][(r+s)^2-(r+s)(3)+(3)^2] \\= [(r+s)+(3)][(r^2+2rs+s^2)-3r-3s+9] \\= (r+s+3)(r^2+2rs+s^2-3r-3s+9) .\end{array}