The equation $x^6 - 3x^5 - 6x^3 - x + 8 = 0$ has
$\textbf{(A)} \text{ no real roots} \\ \textbf{(B)} \text{ exactly two distinct negative roots} \\ \textbf{(C)} \text{ exactly one negative root} \\ \textbf{(D)} \text{ no negative roots, but at least one positive root} \\ \textbf{(E)} \text{ none of these}$
Let $P(x) = x^6 - 3x^5 - 6x^3 - x + 8$. When $x  0$. 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 $\boxed{\textbf{(D)}}$.