1. Given the function $f(x)=-2 x^{3}-x$. If $x_{1} 、 x_{2}$ 、 $x_{3} \in \mathbf{R}$ and $x_{1}+x_{2}>0, x_{2}+x_{3}>0, x_{3}+x_{1}>0$. Then $f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)(\quad)$.
(A) greater than zero
(B) less than zero
(C) equal to zero
(D) greater than zero or less than zero
$-1 . B$.
It is known that $f(x)$ is an odd function and a decreasing function on $\mathbf{R}$, so $f\left(x_{1}\right)<f\left(-x_{2}\right)$,

which means $f\left(x_{1}\right)+f\left(x_{2}\right)<0$.
Similarly, $f\left(x_{2}\right)+f\left(x_{3}\right)<0, f\left(x_{3}\right)+f\left(x_{1}\right)<0$.
Therefore, $f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)<0$.