Answer
$7y\sqrt[3]{2x}$
Work Step by Step
Simplify each radical by factoring the radicand so that at least one factor is a perfect cube to obtain:
$=\sqrt[3]{27y^3(2x)} + y\sqrt[3]{64(2x)}
\\=\sqrt[3]{(3y)^3(2x)} + y\sqrt[3]{(4^3)(2x)}
\\=3y\sqrt[3]{2x} + 4y\sqrt[3]{2x}$
RECALL:
The distributive property states that for any real numbers a, b, and c:
(1) $ac + bc = (a+b)c$
(2) $ac-bc=(a-b)c$
Use rule (1) above to combine like terms and obtain:
$=(3y+4y)\sqrt[3]{2x}
\\=7y\sqrt[3]{2x}$
