jmc

We note that $$x^3-2x^2-8x+4=(x^2-2x-8)\cdot x+4=0\cdot x+4,$$since $x^2-2x-8=0$. Now, $0\cdot x+4=\boxed{4}$, so this is our answer.

We could also solve for $x$ from the information given. The expression $x^2-2x-8$ factors as $(x+2)(x-4)$. Thus $x$ must be equal to 4 or $-2$. Since $x$ is positive, $x$ must equal 4. Thus our expression is equal to $$4^3-2\cdot 4^2-8\cdot 4+4.$$We can factor out a 4 to find that this is $$4(4^2-2\cdot 4-8+1)=4(16-8-8+1)=4\cdot 1=4,$$as before.

(Alternatively, since the problem statement implies that there is only one positive value of $x$ such that $x^2-2x-8=0$, we could find the value 4 by trial and error, and then simplify as above.)