Factorise $64 - (x + 1)^3$. Hence, solve $64 - (x + 1)^3 = 15(3 - x)$, expressing non-integer roots in surd form.

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

$64 - (x + 1)^3 = 4^3 - (x + 1)^3$
$= [4 - (x + 1)][4^2 + 4(x + 1) + (x + 1)^2]$

2. Simplify the expressions inside the brackets

$= (4 - x - 1)[16 + 4x + 4 + (x^2 + 2x + 1)]$
$= (3 - x)(x^2 + 6x + 21)$

1. Substitute the factorised form from part (a) into the equation

From part (a), $64 - (x + 1)^3 = (3 - x)(x^2 + 6x + 21)$.
Substitute this into the equation:
$(3 - x)(x^2 + 6x + 21) = 15(3 - x)$

2. Rearrange and factorise the equation

$(3 - x)(x^2 + 6x + 21) - 15(3 - x) = 0$
$(3 - x)(x^2 + 6x + 21 - 15) = 0$
$(3 - x)(x^2 + 6x + 6) = 0$

3. Solve for the integer root

$3 - x = 0 \implies x = 3$

4. Solve the quadratic factor using the quadratic formula

$x^2 + 6x + 6 = 0$
$x = \frac{-6 \pm \sqrt{6^2 - 4(1)(6)}}{2(1)}$
$x = \frac{-6 \pm \sqrt{36 - 24}}{2}$
$x = \frac{-6 \pm \sqrt{12}}{2}$
$x = \frac{-6 \pm 2\sqrt{3}}{2}$
$x = -3 \pm \sqrt{3}$