$$1=\sqrt[3]{\frac{13\sqrt{3}}{36}+\frac{5}{8}}- \sqrt[3]{\fra... ...}{8}} \qquad 2=\sqrt[3]{6\sqrt{3}+10}- \sqrt[3]{6\sqrt{3}-10}$$

More generally, for arbitrary $c$,

$$1=\sqrt[3]{\frac{1}{8}+\frac{c^2}{2}+\frac{c(4c^2+9)\sqrt{3}... ...qrt[3]{\frac{1}{8}+\frac{c^2}{2}-\frac{c(4c^2+9)\sqrt{3}}{36}}$$