jmc

Since the denominator involves cube roots, we can't just multiply by a conjugate. Instead we use the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$. Letting $a=\sqrt[3]{3}$ and $b=\sqrt[3]{2}$, we have $$\frac{1}{\sqrt[3]{3}-\sqrt[3]{2}}=\frac{1}{\sqrt[3]{3}-\sqrt[3]{2}}\cdot \frac{(\sqrt[3]{3}{)}^2+\sqrt[3]{3}\sqrt[3]{2}+(\sqrt[3]{2}{)}^2}{(\sqrt[3]{3}{)}^2+\sqrt[3]{3}\sqrt[3]{2}+(\sqrt[3]{2}{)}^2}.$$The denominator simplifies by the above identity to $(\sqrt[3]{3}{)}^3-(\sqrt[3]{2}{)}^3=1$, so we are left with $\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}$. Matching the form given in the problem, $D=1$ and $A=9$, $B=6$, $C=4$ (in some order), so $A+B+C+D=\boxed{20}$.