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)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:  x×y×z)+(x+y)×(y+z)×(z+x)=(x+y+z)×(x×y+y×z)+z×x) \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
4 years, 10 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Yes; I really like this identity too. For example it can be used to prove that r1+r2+r3r=4Rr_1 + r_2 + r_3 - r = 4R (from Incircles and Excircles). If we substitute x=sax = s-a etc., then

s(sb)(sc)+s(sc)(sa)+s(sa)(sb)(sa)(sb)(sc)=abcs(s-b)(s-c)+s(s-c)(s-a)+s(s-a)(s-b)-(s-a)(s-b)(s-c) = abc

where ss is the semi-perimeter, and this reduces nicely using area formulas to the desired relationship.

Michael Ng - 4 years, 10 months ago

Log in to reply

@Michael Ng created this problem which uses the identity.

Calvin Lin Staff - 4 years, 9 months ago

Log in to reply

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!

Aareyan Manzoor - 4 years, 9 months ago

Log in to reply

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

Curtis Clement - 4 years, 10 months ago

Log in to reply

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

Atanu Ghosh - 4 years, 1 month ago

Log in to reply

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

Niranjan Khanderia - 1 year, 5 months ago

Log in to reply

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

Lu Chee Ket - 4 years, 10 months ago

Log in to reply

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)

Daniel Liu - 4 years, 10 months ago

Log in to reply

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!

Lu Chee Ket - 4 years, 10 months ago

Log in to reply

Thanks. I did not realize this very useful identity.

Niranjan Khanderia - 4 years, 9 months ago

Log in to reply

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

Frankie Fook - 4 years, 9 months ago

Log in to reply

It's following the rules of principle of duality

Akhil Bansal - 4 years, 1 month ago

Log in to reply

No, that is not the principle of duality.

Calvin Lin Staff - 4 years, 1 month ago

Log in to reply

I mean l'll bit similar to that

Akhil Bansal - 4 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...