Simplify $\sqrt{8+2 \sqrt{10+2 \sqrt{5}}}$ $+\sqrt{8-2 \sqrt{10+2 \sqrt{5}}}$, the value equals ( ).
(A) 4
(B) $2+\sqrt{5}$
(C) $\sqrt{10}+\sqrt{2}$
(D) $2 \sqrt{5}+2$
(1994, Zu Chongzhi Cup Mathematics Competition)
$$
\begin{array}{l}
\text { Sol } \quad \text { Let the original expression }=x, \quad \text { then } x^{2}=8+ \\
2 \sqrt{10+2 \sqrt{5}}+2 \sqrt{64-4(10+2 \sqrt{5})} \\
+8-2 \sqrt{10-2 \sqrt{5}}, \\
x^{2}=16+4(\sqrt{5}-1) \\
\quad=(\sqrt{10}+\sqrt{2})^{2} . \\
\because x>0, \\
\therefore x=\sqrt{10}+\sqrt{2} .
\end{array}
$$

Note. This method is applicable to $\sqrt[n]{A+\sqrt[n]{B}} \pm \sqrt[n]{A-\sqrt[n]{B}}$.
$$