smc

Let $P(x)=x^6-3x^5-6x^3-x+8$. When $x<0$, $P(x)>0$. Therefore, there are no negative roots. Notice that $P(1)=-1$ and $P(0)=8$. There must be at least one positive root between 0 and 1, therefore the answer is $\boxed{\text{ no negative roots, but at least one positive root}}$.