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To solve this related rates problem, let’s start by breaking it down with some definitions and applying the relevant formulas.

### Step 1: Define Variables and Formulas

1. **Volume of a cube** \( V = s^3 \), where \( s \) is the side length.
2. **Surface area of a cube** \( A = 6s^2 \).

### Step 2: Given Information

- The volume of the cube is increasing at a constant rate: \(\frac{dV}{dt} = 1330 \, \text{in}^3/\text{s}\).
- The side length of the cube at the instant we are considering is \( s = 8 \, \text{in} \).
- We need to find the rate of change of the surface area, \(\frac{dA}{dt}\), at the moment when \( s = 8 \, \text{in} \).

### Step 3: Relate the Volume to Side Length

Differentiating \( V = s^3 \) with respect to time \( t \):
\[
\frac{dV}{dt} = 3s^2 \frac{ds}{dt}
\]
Plugging in \( \frac{dV}{dt} = 1330 \) and \( s = 8 \):
\[
1330 = 3(8^2) \frac{ds}{dt}
\]
\[
1330 = 3 \times 64 \times \frac{ds}{dt}
\]
\[
1330 = 192 \frac{ds}{dt}
\]
\[
\frac{ds}{dt} = \frac{1330}{192} \approx 6.927 \, \text{in}/\text{s}
\]

### Step 4: Differentiate the Surface Area with Respect to Time

Now, we differentiate \( A = 6s^2 \) with respect to \( t \):
\[
\frac{dA}{dt} = 12s \frac{ds}{dt}
\]
Substitute \( s = 8 \) and \( \frac{ds}{dt} \approx 6.927 \):
\[
\frac{dA}{dt} = 12 \times 8 \times 6.927
\]
\[
\frac{dA}{dt} \approx 665.792 \, \text{in}^2/\text{s}
\]

### Answer

The rate of change of the surface area of the cube at the given instant is approximately:
\[
\frac{dA}{dt} \approx 665.792 \, \text{in}^2/\text{s}
\]
Rounded to three decimal places, this is \( 665.792 \, \text{in}^2/\text{s} \).

Would you like further clarification on any steps? 

Here are some related questions you may find useful:
1. How does the rate of change in side length affect the rate of change in surface area?
2. What would happen to \(\frac{dA}{dt}\) if the rate \(\frac{dV}{dt}\) was different?
3. How does the surface area formula change if the shape is a rectangular prism instead of a cube?
4. Can you explain why we needed to differentiate twice in this problem?
5. What would be the result if the side length \( s \) was changing at a constant rate instead of the volume?

**Tip:** For related rates problems, always ensure the relationships between the variables are clear before differentiating with respect to time.

The volume of a cube is increasing at a constant rate of 1330 cubic inches per second. At the instant when the side length of the cube is 8 inches, what is the rate of change of the surface area of the cube? Round your answer to three decimal places (if necessary).

Explore how to find the rate of change of the surface area of a cube when its volume increases at 1330 cubic inches per second, given a side length of 8 inches. This related rates problem covers differentiation and the application of volume and surface area formulas for a cube. Suitable for Grade 11-12 students.

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