To solve this, we need to use related rates involving the volume and surface area of a cube. Let $x$ be the length of an edge of the cube (in cm), $V$ the volume, and $A$ the surface area.

Step 1: Relations for Volume and Surface Area

1. The volume of a cube is given by: $V = x^3$
2. The surface area of a cube is given by: $A = 6x^2$

Step 2: Differentiating Volume and Surface Area with Respect to Time

1. Differentiate the volume equation $V = x^3$ with respect to time $t$: $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$ We are given that $\frac{dV}{dt} = 6 \, \text{cm}^3/\text{s}$.

2. Differentiate the surface area equation $A = 6x^2$ with respect to time $t$: $\frac{dA}{dt} = 12x \frac{dx}{dt}$

Step 3: Solve for $\frac{dx}{dt}$

Using the equation from the volume: $6 = 3x^2 \frac{dx}{dt}$ Solve for $\frac{dx}{dt}$: $\frac{dx}{dt} = \frac{6}{3x^2} = \frac{2}{x^2}$

Step 4: Substitute $\frac{dx}{dt}$ into the Surface Area Equation

Now, substitute $\frac{dx}{dt} = \frac{2}{x^2}$ into the equation for $\frac{dA}{dt}$: $\frac{dA}{dt} = 12x \times \frac{2}{x^2} = \frac{24}{x}$

So, the rate at which the surface area is increasing is: $\frac{dA}{dt} = \frac{24}{x} \, \text{cm}^2/\text{s}$

Final Answer:

The surface area is increasing at the rate of $\frac{24}{x} \, \text{cm}^2/\text{s}$ when the edge length is $x$ cm.

Would you like to go over any step in more detail, or explore some variations of this problem?

Here are 5 related questions:

1. What happens if the rate of change of the volume was different?
2. How would the problem change if the shape were a sphere?
3. What is the rate of change of the diagonal of the cube?
4. How fast is the edge length increasing when the surface area is increasing at 10 cm²/s?
5. How does the surface area change when the edge is doubled?

Tip: When solving related rates problems, make sure to differentiate each variable with respect to time carefully!