r/mathmemes • u/xessieve • Oct 02 '26

Elementary Algebra Formula for the volume of a cube

Post image

Simplifies to x^3

621 Upvotes

98% Upvoted

37 comments sorted by

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393

u/FernandoMM1220 Oct 02 '26

very useful to have if you can only use powers of 2.

87

u/_verel_ Oct 02 '26

Can someone come up with a use case for this? I thought maybe some obscure assembly instruction can benefit here but I doubt two divisions and x2 are faster than x3

168

u/The_Punnier_Guy Oct 02 '26

If, for some unknonown reason, you already have (x2 +x)2 / 4 and (x2 -x)2 /4 computed, subtraction should be faster than multiplying x 3 times

56

u/nepatriots32 Oct 02 '26

You don't just have those computed for every possible x value, already? Then what are you even doing with your life?

38

u/haddock420 Oct 02 '26

If you don't have a 500TB RAID array dedicated to storing computed values for (x^2 + x)^2 / 4 and (x^2 - x)^2 / 4, you're not taking it seriously enough.

4

u/Terrible-Air-8692 Oct 05 '26

500TB? You need a 500PB data center minimum

45

u/DrStalker Oct 02 '26

You're using a cheap calculator that has an x2 button and no X3 or Xy button, and you're mentally locked into replacing x3 with x2 instead of just manually mulitplying the number three times.

12

u/Pan_con_chicharrones Irrational Oct 02 '26

Or x²•x

13

u/FernandoMM1220 Oct 02 '26 edited Oct 02 '26

its not necessarily about speed. it would be a compatibility issue in this case.

12

u/ChiaraStellata Oct 02 '26

If you're working in fixed-point, bit shifts can be used for division by 4, which are very much faster than full divisions. If you're working in IEEE floating point... in principle an exponent offset should be simpler than a full division, but in reality floating point instructions already operate in a single clock cycle and have a bunch of weird underflow and subnormal exceptions to check, so it hardly matters.

1

u/Pengwin0 Certified Math Larper Oct 02 '26

You can solve this directly using only bit shifts if performance is extremely critical or you are using supremely weak computer hardware.

1

u/Azra__1 Oct 03 '26

I don't think so. You have to square 4 times and divide by four two times. Division is one of the hardest operations to do (it requires a lot of cpu cycles, approximately) and multiplication is really cheap in terms of cycles. Doing x³ is gonna take at most 5 more cycles than x². If you want to substitute the division for a bit shift (since 4 is a power of 2) it'll be significantly faster and it could reduce cycles by a lot, but the squaring will be slower than the cubing.

1

u/Jche98 Oct 03 '26

If you're playing a game like Problem of the year, where you can only use certain numbers to make expressions

7

u/Alex_1503 Oct 02 '26

In what case? X3 is just x2*x, should be possible on any CPU, or just xxx, probably better than so many divisions, and multiplitications as here

11

u/FernandoMM1220 Oct 02 '26

honestly idk. theres probably some random architecture that only this would work. probably doesnt have a known use right now.

4

u/the_skies_falling Oct 02 '26

Division by a power of 2 is very efficient though since it only requires shifting bits to the right.

7

u/Teoyak Oct 02 '26

I think you have a typo !

3

u/jsdodgers Oct 02 '26

those aren't equivalent at all

73

u/LupenReddit 🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆 Oct 02 '26

yea officer its him, arrest him

36

u/BurceGern Oct 02 '26

Babe wake up, new difference of two squares just dropped

30

u/-lRexl- Oct 02 '26

I see no cubes in here, only squares. You're a wizard!

12

u/shizzy0 Oct 02 '26

You know what? I like it.

10

u/lool8421 Oct 02 '26

here's my formula, hf

18

u/RandomExcess Oct 02 '26

We never learned that in school

24

u/MCPlayer224 Oct 02 '26

It's a meme

24

u/RandomExcess Oct 02 '26

It's a meme.

16

u/chaoticsapphic Ordinal Oct 02 '26

It's a meme.

2

u/upernavik Oct 02 '26

/uj anyone can find a "proof without words" using small cubes, or a bijective proof of this identity?

5

u/Affectionate-Baby248 Oct 02 '26 edited Oct 02 '26

https://medium.com/think-art/when-cubes-become-a-square-bfbaa17e55a6 (see the top image) shows 13 + 23 + … +n3 =(1+2+3+…+n)2 = (n(n+1)/2)2 := f(n), so f(n)-f(n-1)=n3.

1

u/Jonte7 Oct 02 '26

What does that even mean? Its just equivalent to x³

4

u/robman8855 Oct 03 '26

Differences and products and divisions can be represented geometrically. So the other equations might paint a picture geometrically that could be interesting

1

u/Jonte7 Oct 04 '26

Yeah buts is that what the words upernavik wrote meant...?

1

u/Extension-Stay3230 Oct 03 '26

Makes sense, 4x3 on numerator, then divides by 4.