How to factor this expression completely?

Question

$9x^2(4y^2-4y+1)-w^2z^2(4y^2-4y+1)$

I am ending up with $(2y-1)^2 (9x^2-w^2z^2)$. Can I go further? If so, I am unable to see it. Thanks

Answer 1

Here is a hint: the second factor is a difference of squares, namely $$9x^2 - w^2 z^2 = (3x)^2 - (wz)^2.$$

Answer 2

$(9x^2-w^2z^2)$ can further be factorized as $(3x+wz)(3x-wz)$

using the formula $(a^2-b^2)=(a+b)(a-b)$