(i) Calculate \(f(x) = x^3 - 2x - 2\) at \(x = 1\) and \(x = 2\):

\(f(1) = 1^3 - 2 \times 1 - 2 = -3\)

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

Since \(f(1) < 0\) and \(f(2) > 0\), the root lies between \(x = 1\) and \(x = 2\).

(ii) The iterative formula is \(x_{n+1} = \frac{2x_n^3 + 2}{3x_n^2 - 2}\). Rearrange to show it satisfies \(x^3 - 2x - 2 = 0\):

Multiply both sides by \(3x_n^2 - 2\):

\(x_{n+1}(3x_n^2 - 2) = 2x_n^3 + 2\)

Rearrange to \(x_{n+1} = x_n\) to show convergence to the root.

(iii) Use the iterative formula:

Start with \(x_1 = 1.5\):

\(x_2 = \frac{2(1.5)^3 + 2}{3(1.5)^2 - 2} = 1.8421\)

\(x_3 = \frac{2(1.8421)^3 + 2}{3(1.8421)^2 - 2} = 1.7549\)

\(x_4 = \frac{2(1.7549)^3 + 2}{3(1.7549)^2 - 2} = 1.7693\)

\(x_5 = \frac{2(1.7693)^3 + 2}{3(1.7693)^2 - 2} = 1.7669\)

The root correct to 2 decimal places is 1.77.