Answer
$-6x^{5}y^{4}\sqrt[3]{2x^2}$
Work Step by Step
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
\sqrt[3]{-24x^{10}y^8}\sqrt[3]{18x^7y^4}
\\\\=
\sqrt[3]{(-8\cdot3)(2\cdot9)x^{10+7}y^{8+4}}
\\\\=
\sqrt[3]{(-8\cdot27\cdot2)x^{17}y^{12}}
.\end{array}
Extracting the factors that are perfect powers of the index, the given expression simplifies to
\begin{array}{l}\require{cancel}
\sqrt[3]{(-8\cdot27\cdot2)x^{17}y^{12}}
\\\\=
\sqrt[3]{(-8\cdot27)x^{15}y^{12}\cdot2x^2}
\\\\=
\sqrt[3]{[(-2\cdot3)x^{5}y^{4}]^3\cdot2x^2}
\\\\=
\sqrt[3]{[(-2\cdot3)x^{5}y^{4}]^3\cdot2x^2}
\\\\=
(-2\cdot3)x^{5}y^{4}\sqrt[3]{2x^2}
\\\\=
-6x^{5}y^{4}\sqrt[3]{2x^2}
.\end{array}
