jmc

We have that $$\begin{array}{rl}& ((2x^2+3x+3)-(x^2+6x-6))(x+3)\\ & =(x^2-3x+9)(x+3)\\ & =x(x^2-3x+9)+3(x^2-3x+9)\\ & =x^3-3x^2+9x+3x^2-9x+27\\ & =\boxed{x^3+27}.\end{array}$$