Factorise $3x^3 - 24y^3$ completely. Express $\frac{7x^2 + 19x + 15}{(x + 1)^2(x + 2)}$ as partial fractions.

1. Factor out the common numerical factor

$3x^3 - 24y^3 = 3(x^3 - 8y^3)$

2. Apply the difference of cubes identity

Recognise that $8y^3 = (2y)^3$.
Using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ with $a = x$ and $b = 2y$:
$3(x^3 - (2y)^3) = 3(x - 2y)(x^2 + x(2y) + (2y)^2)$
$= 3(x - 2y)(x^2 + 2xy + 4y^2)$

1. Set up the correct form for the partial fractions

Let $\frac{7x^2 + 19x + 15}{(x + 1)^2(x + 2)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x + 2}$

2. Form the basic algebraic identity

$7x^2 + 19x + 15 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)^2$

3. Find the value of $B$ by substituting $x = -1$

Substitute $x = -1$:
$7(-1)^2 + 19(-1) + 15 = B(-1 + 2)$
$7 - 19 + 15 = B(1)$
$B = 3$

4. Find the value of $C$ by substituting $x = -2$

Substitute $x = -2$:
$7(-2)^2 + 19(-2) + 15 = C(-2 + 1)^2$
$28 - 38 + 15 = C(1)$
$C = 5$

5. Find the value of $A$ by comparing coefficients of $x^2$

Compare coefficients of $x^2$:
$7 = A + C$
$7 = A + 5$
$A = 2$

Therefore, $\frac{7x^2 + 19x + 15}{(x + 1)^2(x + 2)} = \frac{2}{x + 1} + \frac{3}{(x + 1)^2} + \frac{5}{x + 2}$