jmc

Since $-2$ and 3 are roots, $$\begin{array}{rl}a(-2{)}^3+(a+2b)(-2{)}^2+(b-3a)(-2)+(8-a)& =0,\\ a(3{)}^3+(a+2b)3^2+(b-3a)(3)+(8-a)& =0.\end{array}$$Solving, we find $a=\frac{8}{9}$ and $b=-\frac{40}{27}.$ By Vieta's formulas, the sum of the roots is $$-\frac{a+2b}{a}=\frac{7}{3},$$so the third root is $\frac{7}{3}-(-2)-3=\boxed{\frac{4}{3}}.$