Example 8 Given real numbers $a \neq b$, and 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
(2004, “TRULY ${ }^{\circledR}$ Xinli Cup” National Junior High School Mathematics Competition)
Given that $a$ and $b$ are the roots of the equation
$$
(x+1)^{2}+3(x+1)-3=0
$$

After rearranging, we get $x^{2}+5 x+1=0$. Since
$$
\Delta=25-4>0, a+b=-5, a b=1,
$$

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

Thus, the correct answer is (B).