## Precalculus (6th Edition) Blitzer

$2$
First, recognize some squares and cubes: $13^{2}=169,$ $3^{2}=9$ $10^{3}=1000$ $6^{3}=216$ $\sqrt[3]{\sqrt{\sqrt{169}+\sqrt{9}}+\sqrt{\sqrt[3]{1000}+\sqrt[3]{216}}}=\sqrt[3]{\sqrt{13+3}+\sqrt{10+6}}$ $=\sqrt[3]{\sqrt{16}+\sqrt{16}}$ $=\sqrt[3]{4+4}$ $=\sqrt[3]{8}$ $=2$