Answer
$\sqrt[3]{12} + 4\sqrt[3]{10}$
Work Step by Step
RECALL:
(1) Distributive Property:
For any real numbers a, b, and c,
$a(b+c)=ab+ac$
(2) For any real numbers real numbers a and b within the domain,
$\sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{ab}$
Use rule (1) above to obtain:
$=\sqrt[3]{2}(\sqrt[3]{6}) + \sqrt[3]{2} \cdot 4\sqrt[3]{5}$
Use rule (2) above to obtain:
$=\sqrt[3]{2(6)} + 4\sqrt[3]{2(5)}
\\=\sqrt[3]{12} + 4\sqrt[3]{10}$
