1. Given real numbers $a \neq b$, and satisfy
$$
(a+1)^{2}=3-3(a+1), 3(b+1)=3-(b+1)^{2} \text {. }
$$

Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is ( ).
(A) 23
(B) -23
(C) -2
(D) -13
$-、 1 . B$.
Since $a, b$ are the roots of the equation with respect to $x$
$$
(x+1)^{2}+3(x+1)-3=0
$$

Rearranging this equation, we get
$$
x^{2}+5 x+1=0 \text {. }
$$

Since $\Delta=25-4>0$, we have
$$
a+b=-5, a b=1 \text {. }
$$

Therefore, $a, b$ are both negative. Thus,
$$
\begin{array}{l}
b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}} \\
=-\frac{(a+b)^{2}-2 a b}{\sqrt{a b}}=-23 .
\end{array}
$$