Find the value

On rearranging the terms

$=x^{2}-4 x z+4 z^{2}-y^{2}$

$=(x)^{2}-2 \times x \times 2 z+(2 z)^{2}-y^{2}$

Using the identity $x^{2}-2 x y+y^{2}=(x-y)^{2}$

$=(x-2 z)^{2}-y^{2}$

Using the identity $p^{2}-q^{2}=(p+q)(p-q)$

$=(x-2 z+y)(x-2 z-y)$

$\therefore x^{2}-y^{2}-4 x z+4 z^{2}=(x-2 z+y)(x-2 z-y)$