The sequence of values given by the iterative formula \(x_{n+1} = \frac{3x_n}{4} + \frac{15}{x_n^3}\), with initial value \(x_1 = 3\), converges to \(\alpha\).

(i) Use this iterative formula to find \(\alpha\) correct to 2 decimal places, giving the result of each iteration to 4 decimal places.

(ii) State an equation satisfied by \(\alpha\) and hence find the exact value of \(\alpha\).