(i) To prove the identity \((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \equiv \sin^3 \theta + \cos^3 \theta\):

Expand the left-hand side (LHS):

\(\sin \theta + \cos \theta - \sin^2 \theta \cos \theta - \sin \theta \cos^2 \theta\)

Rearrange terms:

\(= \sin \theta (1 - \cos^2 \theta) + \cos \theta (1 - \sin^2 \theta)\)

Using \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to:

\(= \sin^3 \theta + \cos^3 \theta\)

Thus, the identity is proven.

(ii) Solve \((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) = 3 \cos^3 \theta\):

From part (i), \(\sin^3 \theta + \cos^3 \theta = 3 \cos^3 \theta\)

Rearrange to get:

\(\sin^3 \theta = 2 \cos^3 \theta\)

Divide both sides by \(\cos^3 \theta\):

\(\tan^3 \theta = 2\)

Taking the cube root gives:

\(\tan \theta = \sqrt[3]{2}\)

Calculate \(\theta\) using \(\tan^{-1}(\sqrt[3]{2})\):

\(\theta = 51.6^\circ\) or \(231.6^\circ\) (considering the periodicity of tangent).