I recently found a very interesting Algebraic Identity: \[xyz+(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)\]

What's so special about it? Note that going from one side of the equality to the other, all products are switched with sums, and all sums are switched with products!

This may be seen a bit easier if I rewrite it as follows: \[ \begin{align*} &\color{white}{)}x\color{red}{\times} y\color{red}{\times }z \color{white}{)}\color{blue}{+} \color{grey}{(}x\color{blue}{+}y\color{grey}{)}\color{red}{\times}\color{grey}{(}y\color{blue}{+}z\color{grey}{)}\color{red}{\times} \color{grey}{(}z\color{blue}{+}x\color{grey}{)}\\ =&\color{grey}{(}x\color{blue}{+}y\color{blue}{+}z\color{grey}{)}\color{red}{\times}\color{grey}{(}x\color{red}{\times} y\hspace{0.9ex}\color{blue}{+}\hspace{0.9ex}y\color{red}{\times }z\color{white}{)}\color{blue}{+}\hspace{0.9ex}z\color{red}{\times }x\color{grey}{)} \end{align*}\]

Cool!

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## Comments

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TopNewestYes; I really like this identity too. For example it can be used to prove that \(r_1 + r_2 + r_3 - r = 4R\) (from Incircles and Excircles). If we substitute \(x = s-a\) etc., then

\[s(s-b)(s-c)+s(s-c)(s-a)+s(s-a)(s-b)-(s-a)(s-b)(s-c) = abc\]

where \(s\) is the semi-perimeter, and this reduces nicely using area formulas to the desired relationship.

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@Michael Ng created this problem which uses the identity.

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thanks @Daniel Liu i used this to solve problems like this i wrote a solution using this identity, and i'm thinking about a problem with this identity, will post soon!

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Typo. \(\color{red}{x}\) is missed in color version, line 5.

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F B U L O U S !!!!!!

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I didn't realise that- thanks. It should inspire some good problems :)

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Most awesome discoveries ever!Thanks,this must help a lot.

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Thanks. I did not realize this very useful identity.

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I didn't examine. Good for making questions. However, it could have been found by people in the past.

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Yea, I'm just saying that I just noticed it. I most likely was not the person who discovered it (as seen by the comment by Michael Ng)

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I had just expanded both sides to compare. Should be correct. Do not feel disappointed by what I guessed. You could be the first person to find this. Congratulation!

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It's following the rules of principle of duality

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No, that is not the principle of duality.

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I mean l'll bit similar to that

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