(i) Sketch the curve \(y = \ln(x + 1)\), which is an increasing curve passing through the origin for \(x \geq 0\). The second curve is \(y = 40 - x^3\), which is a decreasing curve for \(x > 0\). The intersection of these curves indicates the root of the equation \(x^3 + \ln(x + 1) = 40\).

(ii) Calculate \(x^3 + \ln(x + 1) - 40\) at \(x = 3\) and \(x = 4\):

For \(x = 3\), \(3^3 + \ln(4) - 40 = 27 + 1.386 - 40 = -11.614\).

For \(x = 4\), \(4^3 + \ln(5) - 40 = 64 + 1.609 - 40 = 25.609\).

Since the sign changes between \(x = 3\) and \(x = 4\), the root lies between these values.

(iii) Using the iterative formula \(x_{n+1} = \sqrt[3]{40 - \ln(x_n + 1)}\):

Start with \(x_1 = 3.5\).

\(x_2 = \sqrt[3]{40 - \ln(3.5 + 1)} = 3.377\).

Continue iterations until the value stabilizes to 3 decimal places: \(x \approx 3.377\).

(iv) For \((e^y - 1)^3 + y = 40\), solve for \(y\):

\(y = \ln(x + 1) \approx 1.48\).