Factorise completely $9b^{2} - 6ab + a^{2} - x^{2}$. Simplify $\dfrac{5}{2x^{2} - 7x - 4} - \dfrac{8}{4 - x}$.

1. Rewrite the first three terms as a perfect square.

$9b^{2}-6ab+a^{2}-x^{2}=(3b-a)^{2}-x^{2}$

2. Factorise the difference of two squares completely.

$(3b-a)^{2}-x^{2}=(3b-a-x)(3b-a+x)$

1. Factorise the quadratic denominator.

$2x^{2}-7x-4=(2x+1)(x-4)$
So
$\frac{5}{2x^{2}-7x-4}-\frac{8}{4-x}=\frac{5}{(2x+1)(x-4)}-\frac{8}{4-x}.$

2. Rewrite the second denominator so both fractions involve $x-4$.

Since $4-x=-(x-4)$,
$-\frac{8}{4-x}=-\frac{8}{-(x-4)}=\frac{8}{x-4}.$
Hence
$\frac{5}{(2x+1)(x-4)}-\frac{8}{4-x}=\frac{5}{(2x+1)(x-4)}+\frac{8}{x-4}.$

3. Combine the fractions using a common denominator and simplify the numerator.

Common denominator: $(2x+1)(x-4)$.
$\frac{5}{(2x+1)(x-4)}+\frac{8}{x-4}$
$=\frac{5}{(2x+1)(x-4)}+\frac{8(2x+1)}{(2x+1)(x-4)}$
$=\frac{5+8(2x+1)}{(2x+1)(x-4)}$
$=\frac{5+16x+8}{(2x+1)(x-4)}$
$=\frac{16x+13}{(2x+1)(x-4)}.$