Factorise:

We have:

$(l+m)^{2}-4 l m=\left(l^{2}+m^{2}+2 l m\right)-4 l m$

$=l^{2}+m^{2}+2 l m-4 l m$

$=l^{2}+m^{2}-2 l m$

$=(l)^{2}+(m)^{2}-2 \times l \times m$

$=(l-m)^{2}$

$\therefore(l+m)^{2}-4 l m=(l-m)^{2}$