Let \(y = f(x) = (3x + 2)^3 - 5\).

To find \(f^{-1}(x)\), solve for \(x\) in terms of \(y\):

\(y = (3x + 2)^3 - 5\)

Add 5 to both sides: \(y + 5 = (3x + 2)^3\)

Take the cube root: \(\sqrt[3]{y + 5} = 3x + 2\)

Subtract 2: \(\sqrt[3]{y + 5} - 2 = 3x\)

Divide by 3: \(x = \frac{\sqrt[3]{y + 5} - 2}{3}\)

Thus, \(f^{-1}(x) = \frac{\sqrt[3]{x + 5} - 2}{3}\).

The domain of \(f^{-1}\) is the range of \(f\), which is \(x \geq 3\).