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

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: $2ab+a^{2}+b^{2} = (a+b)^{2}$

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  • $\begingroup$ The polynomial is not degree 2. It is degree 4. $\endgroup$ – edm Mar 19 '18 at 14:13
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    $\begingroup$ The presumed answer is homogeneous of degree $3$ while the original polynomial is homogeneous of degree $4$. $\endgroup$ – blueInk Mar 19 '18 at 14:13
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    $\begingroup$ On the other hand, observe that $2ab+a^2+b^2=(a+b)^2$. Therefore, you again have a difference of squares $(a+b)^2-c^2$ in the first factor. Likewise, $2ab-a^2-b^2+c^2=c^2-(a+b)^2$. That is another difference of squares. $\endgroup$ – blueInk Mar 19 '18 at 14:14
  • $\begingroup$ Put $c=a+b,-a-b,-a+b,a-b$, get $0$. finish. $\endgroup$ – Takahiro Waki Mar 19 '18 at 14:58
  • $\begingroup$ @gimusi Thanks, I forgot to do that $\endgroup$ – Nicolas Leskiu Mar 22 '18 at 20:44
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HINT

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

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You are almost there.

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

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

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

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