6. Someone viewed 2008 as a number-filling game: $2 \square \square 8$. Thus, he filled each box with a two-digit number $\overline{a b}$ and $\overline{c d}$, and found that the resulting six-digit number $\overline{2 a b c d 8}$ is exactly a perfect cube. Then $\overline{a b}+\overline{c d}=$ (. ).
(A) 40
(B) 50
(C) 60
(D) 70
6.D.

Let $\overline{2 a b c d 8}=(\overline{x y})^{3}$.
According to the characteristic of the last digit, we get $y=2$, and then determine $\overline{x y}$.
Since $60^{3}=216000, 70^{3}=343000$, it follows that $60<\overline{x y}<70$.
Therefore, $\overline{x y}=62$ only. And $62^{3}=238328$, so $\overline{a b}=38, \overline{c d}=32, \overline{a b}+\overline{c d}=70$.