(i) Evaluate the function \(f(x) = x^3 - 8x - 13\) at \(x = 3\) and \(x = 4\):

\(f(3) = 3^3 - 8 \times 3 - 13 = 27 - 24 - 13 = -10\)

\(f(4) = 4^3 - 8 \times 4 - 13 = 64 - 32 - 13 = 19\)

Since \(f(3) < 0\) and \(f(4) > 0\), the root lies between 3 and 4.

(ii) Use the iterative formula \(x_{n+1} = (8x_n + 13)^{\frac{1}{3}}\):

Start with an initial guess, \(x_0 = 3.5\).

\(x_1 = (8 \times 3.5 + 13)^{\frac{1}{3}} = (41)^{\frac{1}{3}} \approx 3.4482\)

\(x_2 = (8 \times 3.4482 + 13)^{\frac{1}{3}} = (40.5856)^{\frac{1}{3}} \approx 3.4325\)

\(x_3 = (8 \times 3.4325 + 13)^{\frac{1}{3}} = (40.46)^{\frac{1}{3}} \approx 3.4301\)

\(x_4 = (8 \times 3.4301 + 13)^{\frac{1}{3}} = (40.441)^{\frac{1}{3}} \approx 3.4297\)

The root is approximately 3.43 to 2 decimal places.

(i) Start with \(x_1 = 3\).

Calculate \(x_2 = \frac{3 \times 3}{4} + \frac{15}{3^3} = 2.75\).

Calculate \(x_3 = \frac{3 \times 2.75}{4} + \frac{15}{2.75^3} = 2.7816\).

Calculate \(x_4 = \frac{3 \times 2.7816}{4} + \frac{15}{2.7816^3} = 2.7794\).

Calculate \(x_5 = \frac{3 \times 2.7794}{4} + \frac{15}{2.7794^3} = 2.7800\).

Calculate \(x_6 = \frac{3 \times 2.7800}{4} + \frac{15}{2.7800^3} = 2.7798\).

Since the values are converging, \(\alpha \approx 2.78\) to 2 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(x = \frac{3}{4}x + \frac{15}{x^3}\).

Rearrange to find \(x^4 = 60\), so \(\alpha = \sqrt[4]{60}\).

(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.

(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.

(i) Evaluate the cubic at \(x = -1\) and \(x = 0\):

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

\(f(0) = 0^3 + 0 + 1 = 1\)

Since \(f(-1) < 0\) and \(f(0) > 0\), there is a root between \(-1\) and \(0\).

(ii) Given \(x_{n+1} = \frac{2x_n^3 - 1}{3x_n^2 + 1}\), rearrange to:

\(x(3x^2 + 1) = 2x^3 - 1\)

\(3x^3 + x = 2x^3 - 1\)

\(x^3 + x + 1 = 0\)

Thus, if the sequence converges, it converges to the root of \(x^3 + x + 1 = 0\).

(iii) Using the iterative formula with \(x_1 = -0.5\):

\(x_2 = \frac{2(-0.5)^3 - 1}{3(-0.5)^2 + 1} = -0.6667\)

\(x_3 = \frac{2(-0.6667)^3 - 1}{3(-0.6667)^2 + 1} = -0.6825\)

\(x_4 = \frac{2(-0.6825)^3 - 1}{3(-0.6825)^2 + 1} = -0.6849\)

\(x_5 = \frac{2(-0.6849)^3 - 1}{3(-0.6849)^2 + 1} = -0.6855\)

The root correct to 2 decimal places is \(-0.68\).