(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 \(-0.68\).