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Would anyone be able to help with showing how the identity $(a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca) - abc.$ I'm absolutely stuck on this question. Sorry that it's so basic $:P.$

Many thanks in advance.

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    When one these things, your last resouece is to multiply everything out and check if they are the same. – 2017-01-10
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    What's stopping you from just expanding the brackets? – 2017-01-10

2 Answers 2

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

What exactly were you stuck with?

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    Got stuck on converting part 3 to part 4. :/ – 2017-01-10
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    @Etched Then convert from part $4$ to part $3$. – 2017-01-10
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    Ahh of course, much clearer now, thanks – 2017-01-10
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Let $a+b+c=S $. Then, $$(a+b)(b+c)(c+a)=(S-a)(S-b)(S-c) $$ $$ =S^3-S^2(a+b+c)+S(ab+bc+ca)-abc $$ $$ =S(ab+bc+ca)- abc $$ as required. Hope it helps.

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    Your $P$ on the last line is entirely unnecessary. Otherwise, good answer. – 2017-01-10
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    @Arthur Thank you for your help. – 2017-01-10
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    @Rohan you should probably remove the abc = *P* seeing as you removed it from the equation ( and thanks a lot for the different approach to the question) – 2017-01-10