jmc

We "complete" the truncated cone by adding a smaller, similar cone atop the cut, forming a large cone. We don't know the height of the small cone, so call it $x$. Since the small and large cone are similar, we have $x/4=(x+6)/8$; solving yields $x=6$. Hence the small cone has radius 4, height 6, and volume $(1/3)\pi (4^2)(6)=32\pi$ and the large cone has radius 8, height 12, and volume $(1/3)\pi (8^2)(12)=256\pi$. The frustum's volume is the difference of these two volumes, or $256\pi -32\pi =\boxed{224\pi }$ cubic cm.