Simplfying radicals

You can break the original value into $\sqrt[3]{16}\sqrt[3]{x^5y^{13}}.$ We know that $\sqrt[3]{16}=2\sqrt[3]{2}$, because of the power,  $2^4.$ Thus, the answer is $\boxed{2\sqrt[3]{2}\sqrt[3]{x^5y^{13}}}$.

....further   y^13 =  y^3 y^3 y^3 y^3 y     so you can take all of those y^3 's out

   and   x^5 = x^3 x^2       so you can take the x^3 out

2 x y^4 $\sqrt[3]{2x^2 y}$