The correct option is (d).

Observe that the given cubic polynomial is equal to $x\left(2007 x^{2}+2006 x+2005\right)$, therefore, $x=0$ is a solution of the given equation and option (a) is ruled out. Since the equation is cubic and $x=0$ is a solution, option (e) is ruled out. Now, to see if the given equation has one, two, or three solutions, we only need to see if the quadratic equation $2007 x^{2}+2006 x+2005=0$ has no solution, or has one or has two solutions. But the discriminant of this equation is

$$
\begin{aligned}
\Delta & =2006^{2}-4 \times 2007 \times 2005=2006^{2}-4(2006+1)(2006-1) \\
& =2006^{2}-4\left(2006^{2}-1\right)=-3 \times 2006^{2}+4<0
\end{aligned}
$$

so that this equation has no real roots. Thus, the initial equation has a unique real root.

Note: Another way (and simpler) to show that $\Delta<0$ is to observe that $2006<2007$ and $2006<4 \times 2005$, therefore,

$$
2006 \times 2006<4 \times 2005 \times 2007 \text { and } 2006^{2}-4 \times 2005 \times 2007<0
$$
Regarding the equation $2007 x^{3}+2006 x^{2}+2005 x=0$, it is correct to state that:
(a) it has no roots;
(d) it has only one real root;
(b) it has three distinct real roots;
(e) it has three positive roots.
(c) it has two equal roots;