$-\sqrt[3]{4} \times \sqrt[3]{-16}$
An odd root of a negative radicand is always a negative$-\sqrt[3]{4} \times \left( -\sqrt[3]{16} \right)$
Multiplying two negatives equals a positive: $\left( - \right) \times \left( - \right)=\left( + \right)$$\sqrt[3]{4}\sqrt[3]{16}$
Calculate the product$\sqrt[3]{64}$
Evaluate the cube root$4$