jmc

Apply the sum of cubes factorization to the expression $1000x^3+27=(10x{)}^3+3^3$ to obtain $$1000x^3+27=(10x+3)(100x^2-30x+9).$$Thus $a^2+b^2+c^2+d^2+e^2+f^2=0^2+10^2+3^2+100^2+(-30{)}^2+9^2=\boxed{11,090}$. Note that the fundamental theorem of algebra implies that the factorization we have given is unique, since the discriminant $(-30{)}^2-4(100)(9)$ of the quadratic $100x^2-30x+9$ is negative.