(i) To find the points of intersection, set the equations equal: \(9 - x^3 = \frac{8}{x^3}\). Multiply through by \(x^3\) to get \(9x^3 - x^6 = 8\), which simplifies to \(x^6 - 9x^3 + 8 = 0\). Factor this as \((X - 1)(X - 8) = 0\) where \(X = x^3\). Thus, \(x^3 = 1\) or \(x^3 = 8\), giving \(x = 1\) or \(x = 2\). Therefore, \(a = 1\) and \(b = 2\).

(ii) The area between the curves is given by the integral \(\int_{1}^{2} \left( 9 - x^3 - \frac{8}{x^3} \right) \, dx\). Integrate to find \(\left[ 9x - \frac{x^4}{4} + \frac{4}{x^2} \right]_{1}^{2}\). Evaluating this gives \(18 - 4 + 1 - (9 - \frac{1}{4} + 4) = 2\frac{1}{4}\).

(iii) The derivatives are \(\frac{dy}{dx} = -3x^2\) for \(y = 9 - x^3\) and \(\frac{dy}{dx} = -\frac{24}{x^4}\) for \(y = \frac{8}{x^3}\). Set \(-3c^2 = -\frac{24}{c^4}\) to find \(c\). Solving gives \(c^6 = 8\), so \(c = \sqrt[6]{8}\) or approximately \(1.41 \ldots\).