(i) Sketch the graphs of \(y = x^3\) and \(y = 3 - x\). The intersection of these graphs gives the solution to \(x^3 = 3 - x\). The graph of \(y = x^3\) is a cubic curve passing through the origin, and \(y = 3 - x\) is a straight line with a negative slope. They intersect at exactly one point, indicating one real root.

(ii) Given the iterative formula \(x_{n+1} = \frac{2x_n^3 + 3}{3x_n^2 + 1}\), if it converges, then \(x_{n+1} = x_n = x\). Substituting into the formula gives \(x = \frac{2x^3 + 3}{3x^2 + 1}\). Rearranging, we get \(x(3x^2 + 1) = 2x^3 + 3\), which simplifies to \(x^3 = 3 - x\), showing convergence to the root found in part (i).

(iii) Using the iterative formula, start with an initial guess, say \(x_0 = 1\). Calculate successive iterations to 5 decimal places:

\(x_1 = \frac{2(1)^3 + 3}{3(1)^2 + 1} = 1.25\)

\(x_2 = \frac{2(1.25)^3 + 3}{3(1.25)^2 + 1} = 1.21622\)

\(x_3 = \frac{2(1.21622)^3 + 3}{3(1.21622)^2 + 1} = 1.21305\)

\(x_4 = \frac{2(1.21305)^3 + 3}{3(1.21305)^2 + 1} = 1.21300\)

\(x_5 = \frac{2(1.21300)^3 + 3}{3(1.21300)^2 + 1} = 1.21300\)

The root correct to 3 decimal places is 1.213.