$K(E[3])$ is obtained by a succession of quadratic extensions of $K(\sqrt[3]{{{\mathcal{D}}}(E)})$ each unramified away from $3$. Neither $\Q(\sqrt{-3})$ nor $\Q(\zeta_9)$ possesses any such quadratic extension, so for $K=\Q(\sqrt{-3})$ the image of the mod-$3$ representation is either trivial (when $\Delta$ is a cube) or of order $3$, in which case it is conjugate to $\left(\begin{smallmatrix} 1 & *\\ 0 & 1\end{smallmatrix} \right)$.

For the other fields, the residual image is contained in the normaliser of the non-split Cartan subgroup, of order $16$. If $E[3]$ were absolutely irreducible then the residual image could not be contained in the non-split Cartan subgroup itself, since that is absolutely reducible; hence there would be a (non-trivial) quadratic character unramified outside $3$ whose kernel has residual image equal to the non-split Cartan subgroup. However, for each of the eight fields $K$ in question, the only quadratic extension ramified only above $3$ is $K(\zeta_3)$, which cuts out a *different* index $2$ subgroup of the non-split Cartan normaliser, since the non-split Cartan contains matrices of determinant $-1$. (Here we are using the fact that the determinant of the representation is the cyclotomic character.) This contradiction shows that $E[3]$ is not absolutely irreducible.

Finally, we prove the claim when 