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I have found a factorization problem that couldn't be solved even after peer review. I request someone to please factorize this expression:

$ -a^2-b^2-c^2+2ab+2bc+2ac $

I am also giving my failed attempt in case it is helpful in any way: $$ -a^2-b^2-c^2+2ab+2bc+2ac \\ =-a^2-b^2-c^2-a^2-b^2-c^2+a^2+b^2+c^2+2ab+2bc+2ac \\ =-2(a^2+b^2+c^2)+(a+b+c)^2 $$ I couldn't make any further progress from here for factoring out the $ a^2+b^2+c^2 $. I'm factorising this for simplifying a formula I'm currently working on, and it would be great if this expression can be made into the square of another expression.

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    $\begingroup$ What is the goal you want to achieve? $\endgroup$ Commented Jan 13 at 10:18
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    $\begingroup$ It’s not going to be the square of any polynomial with integer coefficients because if you substitute $a=b=c=1$, you get a value of 3 which is not a square. $\endgroup$ Commented Jan 13 at 10:20
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    $\begingroup$ You cannot factorize it since you have not defined what this means for you. For me, for example, we have the factorization $$- ((b - c)^2 + a^2 - 2(b + c)a)$$ $\endgroup$ Commented Jan 13 at 10:22
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    $\begingroup$ What do you mean by "gets factorized completely"? And why should this be possible at all? Consider $a^2+b^2+c^2$ itself, for example. $\endgroup$ Commented Jan 13 at 10:24
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    $\begingroup$ What is an "expression"? For example, $a^2+b^2+c^2=1\cdot(a^2+b^2+c^2)$ is a product of expressions. $\endgroup$ Commented Jan 13 at 10:29

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If we allow square roots, we can do this factorization.

Render the expression as

$4ab-(a^2+b^2+c^2+2ab-2ac-2bc),$

where the expression in parentheses is a squared quantity, to wit

$a^2+b^2+c^2+2ab-2ac-2bc=(a+b-c)^2.$

So we have the difference of squares factorization

$(2\sqrt{ab}+a+b-c)(2\sqrt{ab}-a-b+c).$

Then applying the Binomial Theorem and putting in $c=(\sqrt{c})^2$ gives

$[(\sqrt{a}+\sqrt{b})^2-(\sqrt{c})^2][-(\sqrt{a}-\sqrt{b})^2+(\sqrt{c})^2]$

where each of the bracketed terms can be factored again as a difference of squares. We end with

$(\sqrt{a}+\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c}).$

This factorization, with the exponents in the original terms doubled and thus no square root radicals in the final factors, enters into the derivation of Heron's formula for the area of a triangle with given side lengths.

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    $\begingroup$ Ok that factorisation was clean. The first step was something I could have never thought of. Also the way individual terms are shown as square roots just so happens to perfectly align with the reason I tried to factorise it in the first place. Thanks for the solution $\endgroup$ Commented Jan 13 at 14:46
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    $\begingroup$ +1 very elegant - you seem to have intuited the question behind the question. $\endgroup$ Commented Jan 13 at 15:49
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    $\begingroup$ After assuming $x^2=a^2+b^2-2ab\cos\theta$ in the formula of a general quadrilateral which I claimed to be "new" Earlier (which in case you forgot, is here: math.stackexchange.com/q/5118825/1653550) and expanding the brackets then applying the factorisation method you used here, we arrive at heron's formula. This means that the area of general quadrilateral area formula which I claimed to be original is just a reformulated version of the combination of $\frac{1}{2}ab\sin\theta$ and herons formula. I'm extremely sorry for wasting your on it. My intent wasn't to misguide. $\endgroup$ Commented Jan 14 at 12:06

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