Okay, let's see. I need to figure out the cost of a 3-inch cube made from fake gold bricks. The problem says that the cost of the paint is proportional to the surface area, and the cost of the concrete is proportional to the volume. So the total cost should be a combination of these two, right? 

First, let me recall the formulas for surface area and volume of a cube. A cube has all sides equal, so the surface area (SA) of a cube with side length 's' is 6 times the area of one face. Each face is a square with area s², so SA = 6s². The volume (V) of a cube is s³. 

The problem mentions that the paint cost is proportional to the surface area, so let's denote the paint cost as k * SA, where k is the proportionality constant. Similarly, the concrete cost is proportional to the volume, so let's denote that as m * V, where m is another proportionality constant. Therefore, the total cost C would be C = k * 6s² + m * s³.

Now, we have two data points: a 1-inch cube costs $1.30 and a 2-inch cube costs $6.80. We can use these to set up two equations and solve for k and m.

Let's start with the 1-inch cube. Plugging s = 1 into the cost equation:

C₁ = k * 6*(1)² + m*(1)³ = 6k + m = $1.30.

Then, the 2-inch cube:

C₂ = k * 6*(2)² + m*(2)³ = k * 6*4 + m*8 = 24k + 8m = $6.80.

So now we have a system of two equations:

1) 6k + m = 1.30

2) 24k + 8m = 6.80

I need to solve for k and m. Let's use substitution or elimination. Let's try elimination. Multiply the first equation by 8 to make the coefficients of m the same.

Equation 1 multiplied by 8: 48k + 8m = 10.40

Equation 2: 24k + 8m = 6.80

Now subtract equation 2 from equation 1 multiplied by 8:

(48k + 8m) - (24k + 8m) = 10.40 - 6.80

This simplifies to 24k = 3.60

Therefore, k = 3.60 / 24 = 0.15

So k is $0.15 per square inch. Now plug this back into equation 1 to find m.

6k + m = 1.30

6*(0.15) + m = 1.30

0.90 + m = 1.30

m = 1.30 - 0.90 = 0.40

So m is $0.40 per cubic inch.

Now, with k and m known, we can compute the cost for a 3-inch cube. Let's calculate each part.

First, surface area: 6*(3)² = 6*9 = 54 square inches. So the paint cost is 0.15*54 = $8.10.

Then, volume: (3)³ = 27 cubic inches. The concrete cost is 0.40*27 = $10.80.

Total cost is 8.10 + 10.80 = $18.90.

Wait, let me double-check these calculations to make sure I didn't make a mistake.

Calculating k and m again:

From equation 1: 6k + m = 1.30

From equation 2: 24k + 8m = 6.80

Solving for k:

Multiply equation 1 by 8: 48k + 8m = 10.40

Subtract equation 2: (48k +8m) - (24k +8m) = 10.40 -6.80

24k = 3.60 => k=0.15

Then m =1.30 -6*0.15=1.30-0.90=0.40. Correct.

For s=3:

Surface area: 6*9=54. Paint cost: 0.15*54=8.10

Volume:27. Concrete cost:0.40*27=10.80

Total:8.10+10.80=18.90. Yes, that seems right.

Alternatively, maybe check if the initial setup is correct. The problem says the cost of paint is proportional to surface area, and concrete cost is proportional to volume. So total cost is a sum of these two proportional parts, with their own constants. So yes, the equations set up correctly.

Another way to check: Let's verify with the given data points.

For s=1:

Paint:6*1=6. 6*0.15=0.90

Concrete:1*0.40=0.40

Total:1.30. Correct.

For s=2:

Paint:6*4=24. 24*0.15=3.60

Concrete:8*0.40=3.20

Total:3.60+3.20=6.80. Correct.

So the calculations are consistent with the given data. Therefore, the cost for a 3-inch cube is $18.90.

I think that's it. Seems straightforward once the proportionality is set up. Maybe someone could question if there's a different interpretation, but the problem states that the paint cost is proportional to surface area and concrete cost proportional to volume, so summing them makes sense.
Therefore, the cost of a 3-inch cube would be $\boxed{18.90}$ dollars.
</think>