Expand $((2x^2+3x+3)-(x^2+6x-6))(x+3)$.
To expand this expression, I need to simplify the parentheses and then use the distributive property. First, I can subtract the polynomials inside the first parentheses by combining like terms. I get $(2x^2+3x+3)-(x^2+6x-6) = 2x^2+3x+3-x^2-6x+6 = x^2-3x+9$. Now I have $(x^2-3x+9)(x+3)$. To use the distributive property, I need to multiply each term in the first parentheses by each term in the second parentheses, and then add up the results. I can write this as $x^2(x+3)-3x(x+3)+9(x+3)$. Now I can distribute again for each term. I get $x^3+3x^2-3x^2-9x+9x+27$. Finally, I can combine any like terms. I see that the $3x^2$ and $-3x^2$ terms cancel out, as do the $-9x$ and $9x$ terms. I'm left with $x^3+27$. 