jmc

We recognize $8x^3-27$ as a difference of cubes. We can write $8x^3-27$ as $(2x{)}^3-3^3$. We know that $$a^3-b^3=(a-b)(a^2+ab+b^2).$$Thus, $$(2x{)}^3-3^3=(2x-3)(4x^2+6x+9).$$Therefore, $a+b+c+d+e=2-3+4+6+9=\boxed{18}$.