Answer: 753

Small Hint:

There is no $x^2$ term, so $r + s + t = 0$ and each of $r + s,$ $s + t,$ $t + r$ is the negative of a root

Big Hint:

When $x + y + z = 0,$ the identity $x^3 + y^3 + z^3 = 3xyz$ applies; Vieta gives the product of the roots from the constant term

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 $$\begin{aligned} &(-t)^3 + (-r)^3 \\ &\quad {}+ (-s)^3 \\ &= -(r^3 + s^3 + t^3). \end{aligned}$$

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 $-(-753) = 753.$