Example 1 Given real numbers $a, b (a \neq b)$, and they satisfy
$$
\begin{array}{l}
(a+1)^{2}=3-3(a+1), \\
3(b+1)=3-(b+1)^{2} .
\end{array}
$$

Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is ( ).
(A) 23
(B) -23
(C) -2
(D) -13
【Analysis】Transform the known two equations into
$$
\begin{array}{l}
(a+1)^{2}+3(a+1)-3=0, \\
(b+1)^{2}+3(b+1)-3=0 .
\end{array}
$$

It can be seen that $a$ and $b$ are the two roots of the equation with respect to $x$
$$
(x+1)^{2}+3(x+1)-3=0
$$

Solution: From equation (1) in the analysis, simplifying and rearranging yields
$$
x^{2}+5 x+1=0 \text {. }
$$

Since $\Delta=25-4>0, a+b=-5, a b=1$, therefore, both $a$ and $b$ are negative.
Then $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}=-\frac{a^{2}+b^{2}}{\sqrt{a b}}$
$$
=-\frac{(a+b)^{2}-2 a b}{\sqrt{a b}}=-23 \text {. }
$$

Therefore, the answer is B.