Solution:

The cubic has no $x^2$ term, so $r + s + t = 0$ by Vieta's formulas. Hence $r + s = -t,$ $s + t = -r,$ and $t + r = -s,$ and the desired sum is $$(-t)^3 + (-r)^3 + (-s)^3 = -(r^3 + s^3 + t^3).$$

Whenever $r + s + t = 0,$ the identity $r^3 + s^3 + t^3 - 3rst = (r + s + t)(r^2 + s^2 + t^2 - rs - st - tr)$ gives $r^3 + s^3 + t^3 = 3rst.$ By Vieta's formulas, $rst = -\frac{2008}{8} = -251,$ so $r^3 + s^3 + t^3 = -753,$ and the answer is $-(-753) = 753.$