Answer
$(0,-4)\cup(4,+\infty)$
Work Step by Step
$16x\leq x^{3} \rightarrow x^{3}-16x \geq0 \rightarrow x(x^{2}-16)\geq \rightarrow x(x-4)(x+4)\geq0$
These are zero when x=0, x=4,x=-4. These numbers divide the real line into the intervals $(-\infty,0)(0,-4)(-4,4)(4,+\infty)$
Ussing the test points to determine the sign of each factor in each interval: