## Precalculus (6th Edition)

$\color{blue}{\sqrt[3]{2}}$
RECALL: $\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}$ Use the rule above to obtain: $=\sqrt[3(2)]{4} \\=\sqrt[6]{4}$ Since $4=2^2$, the expression above is equivalent to: $=\sqrt[6]{2^2}$ Use the rule $\sqrt[n]{a^m}=a^{m/n}$ to obtain: $=2^{2/6} \\=2^{1/3}$ Use the rule $a^{1/n} = \sqrt[n]{a}$ to obtain: $=\color{blue}{\sqrt[3]{2}}$