5. Given real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
2005 a^{3}=2006 b^{3}=2007 c^{3}=2008 d^{3} \\
\sqrt[3]{2005 a^{2}+2006 b^{2}+2007 c^{2}+2008 d^{2}} \\
=\sqrt[3]{2005}+\sqrt[3]{2006}+\sqrt[3]{2007}+\sqrt[3]{2008} .
\end{array}
$$

Then the value of $a^{-1}+b^{-1}+c^{-1}+d^{-1}$ is ( ).
(A) 1
(B) 0
(C) -1
(D) $\pm 1$
5.D.

Let $x=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$,
$2005 a^{3}=2006 b^{3}=2007 c^{3}=2008 d^{3}=k^{3}$.
Obviously, $a, b, c, d, k$ have the same sign and are not zero, then
$$
\begin{array}{l}
2005 a^{2}=\frac{k^{3}}{a}, 2006 b^{2}=\frac{k^{3}}{b}, \\
2007 c^{2}=\frac{k^{3}}{c}, 2008 d^{2}=\frac{k^{3}}{d} ; \\
2005=\frac{k^{3}}{a^{3}}, 2006=\frac{k^{3}}{b^{3}}, \\
2007=\frac{k^{3}}{c^{3}}, 2008=\frac{k^{3}}{d^{3}} .
\end{array}
$$

From the second given equation, we get
$$
\begin{array}{l}
\sqrt[3]{\frac{k^{3}}{a}+\frac{k^{3}}{b}+\frac{k^{3}}{c}+\frac{k^{3}}{d}} \\
=\sqrt[3]{\frac{k^{3}}{a^{3}}}+\sqrt[3]{\frac{k^{3}}{b^{3}}}+\sqrt[3]{\frac{k^{3}}{c^{3}}}+\sqrt[3]{\frac{k^{3}}{d^{3}}} \\
\sqrt[3]{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}
\end{array}
$$

Thus, $\sqrt[3]{x}=x$.
Therefore, $x=0, x=-1, x=1$.
Since $a, b, c, d$ have the same sign, then $x \neq 0$.
Hence, $x=a^{-1}+b^{-1}+c^{-1}+d^{-1}= \pm 1$.