The combined volume of the three boxes is 532 cubic inches. Let's call the volume of the largest box V1, the volume of the medium box V2, and the volume of the smallest box V3.
We know that V1 + V2 + V3 = 532. This is because the combined volume of the three boxes is given as 532 cubic inches.
The area of one face of the largest box is 49 square inches. Since the boxes are rectangular, let's call the length, width, and height of the largest box L1, W1, and H1 respectively.
The area of one face of the largest box is equal to the length multiplied by the width, which is L1 * W1. We are given that this area is 49 square inches, so we have L1 * W1 = 49.
The volume of the medium box is 218 cubic inches less than the volume of the largest box. This means that V2 = V1 - 218.
Now, let's solve these equations to find the values of V1, V2, and V3.
From L1 * W1 = 49, we can see that the length and width of the largest box could be any two numbers that multiply to give 49. For example, if the length is 7 inches and the width is 7 inches, the area would be 7 * 7 = 49 square inches.
Using V2 = V1 - 218, we can substitute V1 - 218 in place of V2 in the equation V1 + V2 + V3 = 532. This gives us V1 + (V1 - 218) + V3 = 532.
Simplifying this equation, we have 2V1 + V3 = 750.
Since we don't have enough information to solve for specific values of V1 and V3, there are multiple solutions to this problem.
For example, if we let V1 = 350 and V3 = 100, then V2 = V1 - 218 = 350 - 218 = 132.
Another solution could be V1 = 400, V3 = 100, and V2 = V1 - 218 = 400 - 218 = 182.
In conclusion, the problem allows for multiple solutions based on the given information. The specific values of V1, V2, and V3 depend on the lengths, widths, and heights of the boxes, which are not provided in the question.