jmc

Let the roots of $(x-\sqrt[3]{13})(x-\sqrt[3]{53})(x-\sqrt[3]{103})=0$ be $\alpha ,$ $\beta ,$ and $\gamma .$ Then by Vieta's formulas, $$\begin{array}{rl}r+s+t& =\alpha +\beta +\gamma ,\\ rs+rt+st& =\alpha \beta +\alpha \gamma +\beta \gamma ,\\ rst& =\alpha \beta \gamma +\frac{1}{3}.\end{array}$$We have the factorization $$r^3+s^3+t^3-3rst=(r+s+t)((r+s+t{)}^2-3(rs+rt+st)).$$Thus, from the equations above, $$r^3+s^3+t^3-3rst={\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma .$$Hence, $$\begin{array}{rl}{r}^3+s^3+t^3& ={\alpha }^3+{\beta }^3+{\gamma }^3+3(rst-\alpha \beta \gamma )\\ & =13+53+103+1\\ & =\boxed{170}.\end{array}$$