Factorise $a^3 + b^3$. Show that $\frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} = 1 - \frac{\sin 2x}{2}$. Hence, solve the equation $\frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} = 1 - \sin^2 2x$ for $0 \le x \le \pi$.

1. State the factorised form of the sum of two cubes

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

1. Apply the sum of cubes identity from part (a) to the numerator

$\text{LHS} = \frac{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{\sin x + \cos x}$

2. Cancel the common factor and apply the Pythagorean identity

$= \sin^2 x - \sin x \cos x + \cos^2 x$
$= 1 - \sin x \cos x$

3. Apply the double angle formula for sine to reach the RHS

$= 1 - \frac{2\sin x \cos x}{2}$
$= 1 - \frac{\sin 2x}{2} = \text{RHS}$

1. Substitute the identity from part (b) into the equation.

Using the result from (b), the equation becomes:
$1 - \frac{\sin 2x}{2} = 1 - \sin^2 2x$

2. Rearrange the equation into a quadratic equation in terms of sin 2x.

$\sin^2 2x - \frac{\sin 2x}{2} = 0$
$2\sin^2 2x - \sin 2x = 0$

3. Factorise the quadratic equation to find the possible values of sin 2x.

$\sin 2x (2\sin 2x - 1) = 0$
So, $\sin 2x = 0$ or $\sin 2x = \frac{1}{2}$.

4. Solve sin 2x = 0 for the given domain.

Given $0 \le x \le \pi$, the domain for $2x$ is $0 \le 2x \le 2\pi$.
For $\sin 2x = 0$:
$2x = 0, \pi, 2\pi$
$x = 0, \frac{\pi}{2}, \pi$

5. Solve sin 2x = 1/2 for the given domain.

For $\sin 2x = \frac{1}{2}$:
Basic angle $\alpha = \frac{\pi}{6}$
$2x = \frac{\pi}{6}, \pi - \frac{\pi}{6}$
$2x = \frac{\pi}{6}, \frac{5\pi}{6}$
$x = \frac{\pi}{12}, \frac{5\pi}{12}$