5. Use several small cubes of the same size, with surfaces all white or all red, to form a large cube $A B C D-E F G H$ as shown in Figure 2. If the small cubes used on the diagonals $A G$, $B H$, $C E$, and $D F$ of the large cube are all red, and a total of 41 are used, and the rest of the large cube is made up of small cubes that are all white, then the number of small cubes with all white surfaces used is ( ) pieces.
(A) 688
(B) 959
(C) 1290
(D) 1687
5.C.

For a large cube, the number of small cubes used on each of the four space diagonals is the same, so the total number of small cubes used on the four space diagonals should be a multiple of 4. However, according to the given condition, the four space diagonals actually use 41 small cubes, which is an odd number. This indicates that for the given large cube, when calculating the number of small cubes used on each space diagonal, the small cube at the intersection of the four space diagonals is counted 4 times. Therefore, the number of red small cubes used on each space diagonal should be $(41+3) \div 4=11$.
At this point, there are also 11 small cubes on each edge of the large cube. Therefore, the number of white small cubes used is
$$
11 \times 11 \times 11-41=1290 \text {. }
$$