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$bc(b+c)+ac(c+a)-ab(a+b)$


Edit by @Andreas: This should be:

$bc(b+c)+ac(c-a)-ab(a+b)$


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

I did: $b^{2}c+bc^{2}+ac^{2}+a^{2}c-a^{2}b-ab^{2}$ After this step I can't find a way to continue, could you give me a light on this solution?

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  • $\begingroup$ You multiply out the answer and show it's the same. $\endgroup$ Mar 20, 2018 at 16:14
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    $\begingroup$ @Gaffney no, it's not a prof, I must factor out this polynomial $\endgroup$ Mar 20, 2018 at 16:18
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    $\begingroup$ Expand the "answer", you will see that it is not correct. $\endgroup$
    – CY Aries
    Mar 20, 2018 at 16:18
  • $\begingroup$ @CYAries maybe the book wrote wrong them? $\endgroup$ Mar 20, 2018 at 16:19
  • $\begingroup$ If the answer is correct, $bc(b+c)+ac(c+a)$ must be divisible by $a+b$, but it is not. $\endgroup$
    – CY Aries
    Mar 20, 2018 at 16:21

1 Answer 1

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You can proceed as follows (add and subtract the red b, then regroup):

$$ bc(b+c)+ac(c-a)-ab(a+b)\\ = bc(b+c)+ac(\color{red}b+c-(a+\color{red}b))-ab(a+b)\\ = (bc+ac)(b+c)- (ac+ab)(a+b)\\ = c(a+b)(b+c)- a(b+c)(a+b)\\ = (c-a)(a+b)(b+c) $$

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  • $\begingroup$ The identity (with $b$ replaced by $-b$) is symmetric in $a,b,c$ and appears in several contexts. See id3_4_1_3a in Special Algebraic Identities. $\endgroup$
    – Somos
    Mar 20, 2018 at 18:04
  • $\begingroup$ Mr. Andreas, but how have you found that adding and subtracting $b$ would come to the answer? Cause I would never come to find such idea in my whole life. Thanks by the way. $\endgroup$ Mar 21, 2018 at 14:49
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    $\begingroup$ A bit of guessing, a bit of logic. The latter: you don't want to destroy brackets, but regroup them. So you won't expand the expression into elementary products, but keep brackets. As $(a+b)$ and $(b+c)$ are already there, the guess is to transform $(c-a)$ into a linear combination of these two, and - alas! - adding and subtracting $b$ does it. The rest is then stoical, it may as well not have worked out. $\endgroup$
    – Andreas
    Mar 21, 2018 at 15:13

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