We claim that the ratio of the areas is (3/2)^3=27/8, not just for these lengths, but also for every possible set of side-lengths x, y, and z.
To find the volume of the area of the original suitcase, multiply together x, y, and z to get xyz. The modified suitcase will have all these dimensions multiplied by 1 1/2, or 3/2, for a result of 3x/2*3y/x*3z/2=27xyz/8. The ratio is (27xyz/8)/(xyz)=27/8, as desired.
Note that we made no assumptions about the value of x, y, and z throughout the whole solution! Therefore, we can plug ANY side-lengths, including the problem's set (28, 16, 8) into this problem to achieve the same ratio.