jmc

By Vieta's formulas, $r+s+t=-9.$

Since $r,$ $s,$ $t$ are the roots of $x^3+9x^2-9x-8,$ $$x^3+9x^2-9x-8=(x-r)(x-s)(x-t).$$Substituting $x=r+s+t=-9$ into the polynomial, we get $$\begin{array}{rl}(r+s)(r+t)(s+t)& =(-9{)}^3+9\cdot (-9{)}^2-9\cdot (-9)-8\\ & =\boxed{73}.\end{array}$$