## Precalculus (6th Edition)

$\color{blue}{66}$
RECALL: $(a+b)(a^2-ab+b^2) = a^3+b^3$ Note that: $\sqrt[3]{2^2} = (\sqrt[3]{2})^2$ $16=4^2$ Thus, if we let $a=\sqrt[3]{2}$ and $b=4$ The expression above becomes: $=(a+b)(a^2-ba+b^2)$ which is equivalent to $(a+b)(a^2-ab+b^2)$ Therefore, using the formula in the recall part above, the given expression simplifies to: $=(\sqrt[3]{2})^3+4^3 \\=2+64 \\=\color{blue}{66}$