Answer
$\color{blue}{-28ab^2c^3\sqrt[3]{5}}$
Work Step by Step
Factor each radicand so that at least one factor is a perfect cube, then then bring out the cube root of the perfect cube factors to obtain:
$=2\sqrt[3]{216a^3b^6c^9(5)}-8abc\sqrt[3]{125b^3c^6(5)}
\\=2\sqrt[3]{6^3a^3(b^2)^3(c^3)^3(5)}-8abc\sqrt[3]{5^3b^3(c^2)^3(5)}
\\=2(6ab^2c^3)\sqrt[3]{5}-8abc(5bc^2)\sqrt[3]{5}
\\=12ab^2c^3\sqrt[3]{5} -40ab^2c^3\sqrt[3]{5}$
Combine like terms to obtain:
$=(12ab^2c^3-40ab^2c^3)\sqrt[3]{5}
\\=\color{blue}{-28ab^2c^3\sqrt[3]{5}}$
