G3.4 Given that $f(x)=\left(x^{4}+2 x^{3}+4 x-5\right)^{2004}+2004$. If $f(\sqrt{3}-1)=d$, find the value of $d$.
$$
\begin{array}{l}
x=\sqrt{3}-1 \\
(x+1)^{2}=(\sqrt{3})^{2} \\
x^{2}+2 x-2=0
\end{array}
$$

By division, $x^{4}+2 x^{3}+4 x-5=\left(x^{2}+2 x-2\right)\left(x^{2}+2\right)-1=-1$
$$
\begin{aligned}
d & =f(\sqrt{3}-1) \\
& =f(x) \\
& =\left(x^{4}+2 x^{3}+4 x-5\right)^{2004}+2004 \\
& =(-1)^{2004}+2004=2005
\end{aligned}
$$