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Let $V$ be the volume of the cube, $S$ be the surface area, and $x$ be the length of an edge. We are given that $\frac{dV}{dt} = 6~cm^3/s$ and we want to find $\frac{dS}{dt}$ when $x = 8~cm$.
The volume of a cube is given by $V = x^3$, and the surface area is given by $S = 6x^2$.
Differentiating $V = x^3$ with respect to time $t$, we get: $$ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} $$ We are given $\frac{dV}{dt} = 6$, so we have: $$ 6 = 3x^2 \frac{dx}{dt} $$ $$ \frac{dx}{dt} = \frac{6}{3x^2} = \frac{2}{x^2} $$
Differentiating $S = 6x^2$ with respect to time $t$, we get: $$ \frac{dS}{dt} = 12x \frac{dx}{dt} $$
We found that $\frac{dx}{dt} = \frac{2}{x^2}$. Substituting this into the equation for $\frac{dS}{dt}$, we get: $$ \frac{dS}{dt} = 12x \left(\frac{2}{x^2}\right) = \frac{24}{x} $$ Now, we substitute $x = 8~cm$: $$ \frac{dS}{dt} = \frac{24}{8} = 3 $$
The surface area of the cube is increasing at a rate of $3~cm^2/s$.
Final Answer: 3 cm²/s
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