(i) Evaluate \(x^3 - x^2 - 6\) for \(x = 2\) and \(x = 3\):

For \(x = 2\), \(2^3 - 2^2 - 6 = 8 - 4 - 6 = -2\).

For \(x = 3\), \(3^3 - 3^2 - 6 = 27 - 9 - 6 = 12\).

Since the sign changes between \(x = 2\) and \(x = 3\), \(\alpha\) lies between 2 and 3.

(ii) The iterative formula is \(x_{n+1} = \sqrt{x_n + \frac{6}{x_n}}\).

If the sequence converges, then \(x_{n+1} = x_n = \alpha\).

Substitute into the formula: \(\alpha = \sqrt{\alpha + \frac{6}{\alpha}}\).

Square both sides: \(\alpha^2 = \alpha + \frac{6}{\alpha}\).

Multiply through by \(\alpha\): \(\alpha^3 = \alpha^2 + 6\).

Rearrange to \(\alpha^3 - \alpha^2 - 6 = 0\), which is the original equation, confirming convergence to \(\alpha\).

(iii) Use the iterative formula starting with an initial guess, say \(x_0 = 2.5\):

\(x_1 = \sqrt{2.5 + \frac{6}{2.5}} = 2.31662\)

\(x_2 = \sqrt{2.31662 + \frac{6}{2.31662}} = 2.22192\)

\(x_3 = \sqrt{2.22192 + \frac{6}{2.22192}} = 2.21936\)

\(x_4 = \sqrt{2.21936 + \frac{6}{2.21936}} = 2.21900\)

\(x_5 = \sqrt{2.21900 + \frac{6}{2.21900}} = 2.21900\)

The value converges to \(\alpha = 2.219\) correct to 3 decimal places.