jmc

We notice that the terms in the numerator of our fraction are both perfect cubes, suggesting that we can use a sum of cubes factorization. Thus we have $$\begin{array}{rl}\frac{8x^3+27y^3}{2x+3y}& =\frac{(2x{)}^3+(3y{)}^3}{2x+3y}\\ & =\frac{(2x+3y)((2x{)}^2-(2x)(3y)+(3y{)}^2)}{2x+3y}\\ & =(2x{)}^2-(2x)(3y)+(3y{)}^2\\ & =4x^2-6xy+9y^2.\end{array}$$ Thus, $a=4$, $b=-6$, and $c=9$, so $a+b+c=4+(-6)+9=\boxed{7}$.