Factoring 2° degree to 1° degree polynomials

Question

$4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}$

There is a difference o squares but after factoring it, I can't find a way to get to the answer: $(c-a+b)(c+a-b)(a+b-c)$

What I Did: $(2ab)^{2}-(a^{2}+b^{2}-c^{2})^{2}$

$(2ab+a^{2}+b^{2}-c^{2})( 2ab-a^{2}-b^{2}+c^{2})$

After this step I couldn't find any idea. Could you give me a hint please?

Edit: I found something: $2ab+a^{2}+b^{2} = (a+b)^{2}$

Answer 1

HINT

$$(2ab+a^{2}+b^{2}-c^{2})( 2ab-a^{2}-b^{2}+c^{2})=[(a+b)^2-c^{2}][c^2-(a-b)^2]$$

Answer 2

You are almost there.

Note that $$(2ab+a^{2}+b^{2}-c^{2})( 2ab-a^{2}-b^{2}+c^{2})$$

$$((a+b)^2-c^2)(c^2-(a-b)^2)$$

$$ (a+b-c)(a+b+c)(c-a+b)(c+a-b)$$