Factorise $a^3 - 2a^2b - 4a + 8b.$ Solve $\dfrac{3x+2}{9x} = \dfrac{1}{7x-3}.$

1. Group terms so that common factors can be extracted.

$a^3-2a^2b-4a+8b$

$=a^2(a-2b)-4(a-2b)$

2. Extract the common binomial factor.

$a^2(a-2b)-4(a-2b)$

$=(a-2b)(a^2-4)$

1. State the restrictions and clear the denominators by cross-multiplication.

Since the denominators cannot be zero,
$x\neq 0$ and $7x-3\neq 0$, so $x\neq \dfrac{3}{7}$.

$\frac{3x+2}{9x}=\frac{1}{7x-3}$

$(3x+2)(7x-3)=9x$

2. Expand and rearrange into standard quadratic form.

$(3x+2)(7x-3)=9x$

$21x^2+5x-6=9x$

$21x^2-4x-6=0$

3. Factorise the quadratic equation and solve for $x$.

$21x^2-4x-6=0$

$(3x+2)(7x-3)=0$

So
$3x+2=0 \quad \text{or} \quad 7x-3=0$

$x=-\frac{2}{3} \quad \text{or} \quad x=\frac{3}{7}$

But $x=\frac{3}{7}$ is not allowed since it makes $7x-3=0$.

Therefore, the valid solution is
$x=-\frac{2}{3}$