Example 7 Select several colors from the given 6 different colors, to color the six faces of a cube, each face is colored with one color, and two faces sharing a common edge must be of different colors, then the number of different coloring schemes is $\qquad$. (It is agreed that coloring schemes that can coincide after rolling or rotating are considered the same coloring scheme)
Given that three faces sharing a common vertex must be of different colors, at least 3 colors are required. Below, we divide the problem into 4 cases.

(1) When all 6 colors are used, we fix one color on the top face. There are $C_{5}^{1}$ ways to choose 1 color from the remaining 5 colors to color the bottom face. The remaining 4 colors are used to color the 4 side faces (considering the circular permutation of 4 colors), which has $(4-1)!$ ways. Therefore, when all 6 colors are used, there are $C_{5}^{1} \cdot (4-1)! = 30$ coloring schemes.

(2) When only 5 colors are used, there are $C_{6}^{5}$ ways to choose 5 colors from 6 colors. In this case, there must be one pair of opposite faces with the same color. There are $C_{5}^{1}$ ways to choose 1 color from the 5 colors to color this pair of opposite faces, and we place them on the top and bottom. The remaining 4 colors are used to color the 4 side faces (considering the necklace of 4 different colored beads), which has $\frac{1}{2}(4-1)!$ ways. Therefore, when only 5 colors are used, there are $C_{6}^{5} C_{5}^{1} \cdot \frac{1}{2}(4-1)! = 90$ coloring schemes.

(3) When only 4 colors are used, there are $C_{6}^{4}$ ways to choose 4 colors from 6 colors. In this case, there must be 2 pairs of opposite faces with the same color, and one pair of opposite faces with different colors. We place the pair of different colors on the top and bottom, and there are $C_{4}^{2}$ ways to choose 2 colors from the 4 colors to color the top and bottom faces (note that the top and bottom faces are indistinguishable). The remaining 2 colors are used to color the 4 side faces such that the 2 pairs of opposite faces have the same color (considering the necklace of 2 colored beads), which has only 1 way. Therefore, when only 4 colors are used, there are $C_{6}^{4} C_{4}^{2} \cdot 1 = 90$ coloring schemes.

(4) When only 3 colors are used, there are $C_{6}^{3}$ ways to choose 3 colors from 6 colors. In this case, all 3 pairs of opposite faces have the same color, and there is only 1 way to color them with 3 colors. Therefore, when only 3 colors are used, there are $C_{6}^{3} = 20$ coloring schemes.

In summary, the total number of different coloring schemes is $30 + 90 + 90 + 20 = 230$.