Factorise the following:
$\left(2 x+\frac{1}{3}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}$

We use the algebraic identity $a^{2}-b^{2}=(a+b)(a-b)$,where $a = \left(2x + \frac{1}{3}\right)$ and $b = \left(x - \frac{1}{2}\right)$.
Substituting these into the identity:
$= \left[\left(2x + \frac{1}{3}\right) + \left(x - \frac{1}{2}\right)\right] \left[\left(2x + \frac{1}{3}\right) - \left(x - \frac{1}{2}\right)\right]$
Simplify the terms inside the brackets:
$= \left(2x + x + \frac{1}{3} - \frac{1}{2}\right) \left(2x - x + \frac{1}{3} + \frac{1}{2}\right)$
$= \left(3x + \frac{2-3}{6}\right) \left(x + \frac{2+3}{6}\right)$
$= \left(3x - \frac{1}{6}\right) \left(x + \frac{5}{6}\right)$