jmc

Let $r$ be the root which is the sum of the other two. Then $2r$ is the sum of all three roots. But by Vieta's formulas, the sum of the roots of the equation is $-8,$ so we have $2r=-8,$ or $r=-4.$ Thus, $-4$ is a root of the equation, so setting $x=-4,$ we have $$(-4{)}^3+8(-4{)}^2-4(-4)+c=0.$$Solving for $c$ gives $c=\boxed{-80}.$