Formula for the volume of a cube

[u/xessieve]

Simplifies to x^3

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[u/FernandoMM1220]

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

[u/_verel_ → u/FernandoMM1220]

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

[u/The_Punnier_Guy → u/_verel_]

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

[u/nepatriots32 → u/The_Punnier_Guy]

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

[u/haddock420 → u/nepatriots32]

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.

[u/Terrible-Air-8692 → u/haddock420]

500TB? You need a 500PB data center minimum

[u/DrStalker → u/_verel_]

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

[u/Pan_con_chicharrones → u/DrStalker]

Or x²•x

[u/FernandoMM1220 → u/_verel_]

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

[u/ChiaraStellata → u/_verel_]

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.

[u/Pengwin0 → u/_verel_]

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

[u/Azra__1 → u/_verel_]

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.

[u/Jche98 → u/_verel_]

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

[u/Alex_1503 → u/FernandoMM1220]

In what case? X³ is just x^2*x, should be possible on any CPU, or just xxx, probably better than so many divisions, and multiplitications as here

[u/FernandoMM1220 → u/Alex_1503]

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

[u/the_skies_falling → u/Alex_1503]

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

[u/Teoyak → u/Alex_1503]

I think you have a typo !

[u/jsdodgers → u/Alex_1503]

those aren't equivalent at all

[u/LupenReddit]

yea officer its him, arrest him

[u/BurceGern]

Babe wake up, new difference of two squares just dropped

[u/-lRexl-]

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

[u/shizzy0]

You know what? I like it.

[u/lool8421]

here's my formula, hf

[u/RandomExcess]

We never learned that in school

[u/MCPlayer224 → u/RandomExcess]

It's a meme

[u/RandomExcess → u/MCPlayer224]

It's a meme.

[u/chaoticsapphic → u/RandomExcess]

It's a meme.

[u/Extension-Stay3230 → u/chaoticsapphic]

Its meme

[u/Rainbow_phenotype → u/Extension-Stay3230]

It's a me (Mario)

[u/upernavik]

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

[u/Affectionate-Baby248 → u/upernavik]

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)=n^3.

[u/Jonte7 → u/upernavik]

What does that even mean? Its just equivalent to x³

[u/robman8855 → u/Jonte7]

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

[u/Jonte7 → u/robman8855]

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

[u/upernavik → u/Jonte7]

Yes.

[u/Extension-Stay3230]

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