(i) To show that \(\alpha\) lies between 1 and 2, evaluate the function \(f(x) = x^3 - x - 3\) at \(x = 1\) and \(x = 2\):

\(f(1) = 1^3 - 1 - 3 = -3\)

\(f(2) = 2^3 - 2 - 3 = 3\)

Since \(f(1) < 0\) and \(f(2) > 0\), by the Intermediate Value Theorem, there is a root \(\alpha\) between 1 and 2.

(ii) Apply the iterative formulae:

Using formula (A):

\(x_1 = 1.5\)

\(x_2 = (1.5)^3 - 3 = 0.375\)

\(x_3 = (0.375)^3 - 3 = -2.9473\)

The sequence diverges.

Using formula (B):

\(x_1 = 1.5\)

\(x_2 = (1.5 + 3)^{\frac{1}{3}} = 1.6509\)

\(x_3 = (1.6509 + 3)^{\frac{1}{3}} = 1.6684\)

\(x_4 = (1.6684 + 3)^{\frac{1}{3}} = 1.6715\)

\(x_5 = (1.6715 + 3)^{\frac{1}{3}} = 1.6719\)

The sequence converges to \(\alpha = 1.67\) correct to 2 decimal places.