Factorise $\sin^3 x + \cos^3 x$ completely. Show that $\frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} = 1 - \frac{1}{2}\sin 2x$. Hence solve the equation $\frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} = \frac{5}{4}$ for $0^\circ \le x \le 360^\circ$.

1. Apply the sum of cubes identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$

$\sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)$

2. Simplify the expression using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$

$= (\sin x + \cos x)(1 - \sin x \cos x)$

1. Substitute the factorised form of the numerator from part (a) and cancel the common factor.

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

2. Apply the Pythagorean identity to simplify the expression.

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

3. Apply the double angle formula for sine to reach the required form.

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

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

Using the result from (b), the equation becomes:
$1 - \frac{1}{2}\sin 2x = \frac{5}{4}$

2. Isolate $\sin 2x$

$-\frac{1}{2}\sin 2x = \frac{5}{4} - 1$
$-\frac{1}{2}\sin 2x = \frac{1}{4}$
$\sin 2x = -\frac{1}{2}$

3. Find the reference angle and determine the range for $2x$

Basic angle $\alpha = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$
Since $0^\circ \le x \le 360^\circ$, the range for $2x$ is $0^\circ \le 2x \le 720^\circ$.

4. Find all possible values for $2x$ and solve for $x$

Since $\sin 2x$ is negative, $2x$ lies in the 3rd and 4th quadrants.
$2x = 180^\circ + 30^\circ, 360^\circ - 30^\circ, 540^\circ + 30^\circ, 720^\circ - 30^\circ$
$2x = 210^\circ, 330^\circ, 570^\circ, 690^\circ$
$x = 105^\circ, 165^\circ, 285^\circ, 345^\circ$