A metal cube dissolves in acid such that an edge of the cube decreases by $0.60 \mathrm{~mm} / \mathrm{min}$. How fast is the volume of the cube changing when the edge is $8.70 \mathrm{~mm}$ ?
The volume is changing at a rate of $\square \mathrm{mm}^{3} / \mathrm{min}$.
(Round to the nearest integer as needed.)

Step 1 ：This is a problem of related rates. We know that the volume \(V\) of a cube with edge length \(x\) is given by \(V = x^3\). We are given that \(\frac{dx}{dt} = -0.60 \mathrm{~mm} / \mathrm{min}\) (the negative sign indicates that the edge length is decreasing), and we want to find \(\frac{dV}{dt}\) when \(x = 8.70 \mathrm{~mm}\).

Step 2 ：We can find \(\frac{dV}{dt}\) by differentiating \(V = x^3\) with respect to \(t\) using the chain rule, which gives \(\frac{dV}{dt} = 3x^2 \frac{dx}{dt}\). We can then substitute the given values into this equation to find \(\frac{dV}{dt}\).

Step 3 ：Substituting the given values, we get \(\frac{dV}{dt} = 3*(8.70)^2*(-0.60) = -136.242000000000 \mathrm{mm}^{3} / \mathrm{min}\).

Step 4 ：However, the question asks for the answer to be rounded to the nearest integer. So, rounding off, we get \(\frac{dV}{dt} = -136 \mathrm{mm}^{3} / \mathrm{min}\).

Step 5 ：Final Answer: The volume is changing at a rate of \(\boxed{-136} \mathrm{mm}^{3} / \mathrm{min}\).