jmc

Squaring both sides, we get $$9\sqrt[3]{5}-9\sqrt[3]{4}=\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}+2\sqrt[3]{ab}-2\sqrt[3]{ac}-2\sqrt[3]{bc}.$$To make the right side look like the left-side, some terms will probably have to cancel.

Suppose $\sqrt[3]{a^2}=2\sqrt[3]{bc}.$ Then $a^2=8bc,$ so $c=\frac{a^2}{8b}.$ Substituting, the right-hand side becomes $$\begin{array}{rl}\sqrt[3]{b^2}+\sqrt[3]{\frac{a^4}{64b^2}}+2\sqrt[3]{ab}-2\sqrt[3]{a\cdot \frac{a^2}{8b}}& =\sqrt[3]{b^2}+\frac{a}{4b}\sqrt[3]{ab}+2\sqrt[3]{ab}-\frac{a}{b}\sqrt[3]{b^2}\\ & =\left(1-\frac{a}{b}\right)\sqrt[3]{b^2}+\left(\frac{a}{4b}+2\right)\sqrt[3]{ab}.\end{array}$$At this point, we could try to be systematic, but it's easier to test some small values. For example, we could try taking $b=2,$ to capture the $\sqrt[3]{4}$ term. This gives us $$\left(1-\frac{a}{2}\right)\sqrt[3]{4}+\left(\frac{a}{8}+2\right)\sqrt[3]{2a}.$$Then taking $a=20$ gives us exactly what we want: $$\left(1-\frac{20}{2}\right)\sqrt[3]{4}+\left(\frac{20}{8}+2\right)\sqrt[3]{40}=9\sqrt[3]{5}-9\sqrt[3]{4}.$$Then $c=\frac{a^2}{8b}=25.$ Thus, $a+b+c=20+2+25=\boxed{47}.$