Example 3 Choose several colors from the given 6 different colors to color the 6 faces of a cube, with each face being colored with exactly 1 color, and any 2 faces sharing a common edge must be colored differently. Then the number of different coloring schemes is $\qquad$ (Note: If we color two identical cubes and can rotate one of them so that the 6 corresponding faces (up, down, left, right, front, back) of the two cubes have the same color, then we say that the two cubes have the same coloring scheme).
Solution: This problem is relatively complex and requires a classification discussion based on the number of colors used.
(1) If only 3 colors are used, there are $\mathrm{C}_{6}^{3}$ ways to choose 3 colors from 6 different colors. Since every 2 faces sharing a common edge are painted in different colors, the opposite faces of the cube are the same color. Due to the symmetry of the cube, there is only 1 such coloring scheme. Therefore, there are $\mathrm{C}_{6}^{3}=20$ different coloring schemes.
(2) If only 4 colors are used, there are $C_{6}^{4}$ ways to choose 4 colors from 6 different colors. Then, only 1 pair of opposite faces are different colors, with $\mathrm{C}_{4}^{2}$ different coloring methods. Therefore, there are $C_{6}^{4} C_{4}^{2}=90$ different coloring schemes.
(3) If only 5 colors are used, there are $\mathrm{C}_{6}^{5}$ ways to choose 5 colors from 6 different colors. Then, only 1 pair of opposite faces are the same color. Assume these are the top and bottom faces, with $\mathrm{C}_{5}^{1}$ coloring schemes. Then, color the side faces, which have 3 coloring schemes. Therefore, there are $\mathrm{C}_{6}^{5} \mathrm{C}_{5}^{1} \times 3=90$ different coloring schemes.
(4) If 6 different colors are used, then all 6 faces have different colors. Imagine the colors are already painted, and through appropriate rotations, the top face can be made the same color (e.g., red). Then consider the bottom face, which must have 5 different colors. For the bottom face being the same color (e.g., blue), use the remaining 4 colors to paint the side faces, with $\frac{4!}{4}=3$! coloring schemes. Therefore, there are $5 \times 3!=30$ different coloring schemes.

In summary, there are a total of $20+90+90+30=230$ different coloring schemes.