$2 \cdot 34$ When $a<-1$, the roots of the equation
$$
\left(a^{3}+1\right) x^{2}+\left(a^{2}+1\right) x-(a+1)=0
$$

are
(A) two negative roots.
(B) one positive root, one negative root and the absolute value of the negative root is larger.
(C) one positive root, one negative root and the absolute value of the negative root is smaller.
(D) no real roots.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] If $a0, a+10$, we know the equation has unequal real roots. If we assume the two roots of the equation are $x_{1}, x_{2}$, by Vieta's formulas,
$x_{1} x_{2}=-\frac{a+1}{a^{3}+1}0$, which means the absolute value of the negative root is less than the positive root.
Therefore, the answer is $(C)$.