(i) To find \(\alpha\), substitute \(x = -2\) into the equation \(y = x^4 + 2x^3 + 2x^2 - 4x - 16\):

\((-2)^4 + 2(-2)^3 + 2(-2)^2 - 4(-2) - 16 = 16 - 16 + 8 + 8 - 16 = 0\).

Thus, \(\alpha = -2\).

(ii) To show \(\beta\) satisfies \(x = \sqrt[3]{8 - 2x}\), factor the quartic equation:

\((x + 2)(x^3 + 2x - 8) = 0\).

Rearrange \(x^3 + 2x - 8 = 0\) to \(x = \sqrt[3]{8 - 2x}\).

(iii) Use the iterative formula \(x_{n+1} = \sqrt[3]{8 - 2x_n}\) starting with an initial guess, say \(x_0 = 1.5\):

\(x_1 = \sqrt[3]{8 - 2(1.5)} = 1.709975947\).

\(x_2 = \sqrt[3]{8 - 2(1.709975947)} = 1.671585\).

\(x_3 = \sqrt[3]{8 - 2(1.671585)} = 1.667680\).

\(x_4 = \sqrt[3]{8 - 2(1.667680)} = 1.667056\).

Thus, \(\beta \approx 1.67\) to 2 decimal places.