(i) Expand the left-hand side: \((\sin x + \cos x)(1 - \sin x \cos x) = \sin x + \cos x - \sin^2 x \cos x - \cos^2 x \sin x\).

Rearrange terms: \(= \sin x + \cos x - \sin x \cos^2 x - \cos x \sin^2 x\).

Use identities: \(\sin^2 x = 1 - \cos^2 x\) and \(\cos^2 x = 1 - \sin^2 x\).

Substitute: \(\sin x \cos^2 x = \sin x (1 - \sin^2 x) = \sin x - \sin^3 x\) and \(\cos x \sin^2 x = \cos x (1 - \cos^2 x) = \cos x - \cos^3 x\).

Combine: \(\sin x + \cos x - (\sin x - \sin^3 x) - (\cos x - \cos^3 x) = \sin^3 x + \cos^3 x\).

Thus, the identity is proven.

(ii) Given equation: \((\sin x + \cos x)(1 - \sin x \cos x) = 9 \sin^3 x\).

Use part (i): \(\sin^3 x + \cos^3 x = 9 \sin^3 x\).

Rearrange: \(\cos^3 x = 8 \sin^3 x\).

Divide by \(\cos^3 x\): \(\tan^3 x = \frac{1}{8}\).

Take cube root: \(\tan x = \frac{1}{2}\).

Find solutions: \(x = \tan^{-1}(\frac{1}{2}) \approx 26.6^\circ\).

Use periodicity: \(x = 26.6^\circ + 180^\circ = 206.6^\circ\).

Thus, the solutions are \(x = 26.6^\circ\) and \(x = 206.6^\circ\).