# A Most Curious Algebraic Identity

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{aligned} &{~}\color{#D61F06}x \color{#D61F06}{\times} y\color{#D61F06}{\times }z \color{#FFFFFF}{)}\color{#3D99F6}{+} \color{grey}{(}x\color{#3D99F6}{+}y\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times} \color{grey}{(}z\color{#3D99F6}{+}x\color{grey}{)}\\ =&\color{grey}{(}x\color{#3D99F6}{+}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}x\color{#D61F06}{\times} y\hspace{0.9ex}\color{#3D99F6}{+}\hspace{0.9ex}y\color{#D61F06}{\times }z\color{#FFFFFF}{)}\color{#3D99F6}{+}\hspace{0.9ex}z\color{#D61F06}{\times }x\color{grey}{)} \end{aligned}

Cool!

Note by Daniel Liu
6 years, 3 months ago

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Typo. $\color{#D61F06}{x}$ is missed in color version, line 5.

- 2 years, 11 months ago

It's following the rules of principle of duality

- 5 years, 7 months ago

No, that is not the principle of duality.

Staff - 5 years, 7 months ago

I mean l'll bit similar to that

- 5 years, 7 months ago

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

- 5 years, 7 months ago

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

- 6 years, 2 months ago

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!

- 6 years, 3 months ago

Thanks. I did not realize this very useful identity.

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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)

- 6 years, 3 months ago

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!

- 6 years, 3 months ago

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

- 6 years, 3 months ago

Yes; 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.

- 6 years, 3 months ago

@Michael Ng created this problem which uses the identity.

Staff - 6 years, 3 months ago

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