Answer: (a) \(2.50\text{ cm s}^{-1}\); (b) \(240\text{ cm}^2\text{ s}^{-1}\).

For a cube of side \(x\),

\(V=x^3.\)

When \(V=512\),

\(x^3=512,\)

so

\(x=8.\)

(a) Differentiate \(V=x^3\) with respect to \(x\):

\(\frac{dV}{dx}=3x^2.\)

Using the chain rule,

\(\frac{dV}{dt}=\frac{dV}{dx}\frac{dx}{dt}.\)

So at \(x=8\),

\(480=3(8^2)\frac{dx}{dt}.\)

Thus

\(\frac{dx}{dt}=\frac{480}{192}=2.5.\)

The rate of increase of \(x\) is

\(2.50\text{ cm s}^{-1}.\)

(b) The surface area is

\(S=6x^2.\)

Hence

\(\frac{dS}{dx}=12x.\)

Using the chain rule,

\(\frac{dS}{dt}=\frac{dS}{dx}\frac{dx}{dt}.\)

At \(x=8\),

\(\frac{dS}{dt}=12(8)(2.5)=240.\)

So the rate of increase of \(S\) is

\(240\text{ cm}^2\text{ s}^{-1}.\)