S. Ryley bewies 1825 folgendes Theorem:

Jede rationale Zahl kann als Summe von drei rationalen Würfeln ausgedrückt werden.

Herausforderung

Gegeben einige rationale Zahl $r \in \mathbb Q$ drei rationalen Zahlen finden $a,b,c \in \mathbb Q$ , so dass

$$r= a^3+b^3+c^3.$$

Einzelheiten

Ihr Beitrag sollte in der Lage sein, für jede Eingabe eine Lösung zu berechnen, wenn genügend Zeit und Speicher vorhanden sind. Das bedeutet, dass beispielsweise zwei 32-Bit-Zeichen, intdie einen Bruch darstellen, nicht ausreichen.

Beispiele

$$\begin{align} 30 &= 3982933876681^3 - 636600549515^3 - 3977505554546^3 \\ 52 &= 60702901317^3 + 23961292454^3 - 61922712865^3 \\ \frac{307}{1728} &= \left(\frac12\right)^3 + \left(\frac13\right)^3 + \left(\frac14\right)^3 \\ 0 &= 0^3 + 0^3 + 0^3 \\ 1 &= \left(\frac12\right)^3 + \left(\frac23\right)^3 + \left(\frac56\right)^3\\ 42 &= \left(\frac{1810423}{509232}\right)^3 + \left(\frac{-14952}{10609}\right)^3 + \left(\frac{-2545}{4944}\right)^3 \end{align}$$

Pari / GP , 40 Bytes

r->[x=27*r^3+1,9*r-x,z=9*r-27*r^2]/(3-z)

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Gleiche Länge, gleiche Formel:

r->d=9*r^2-3*r+1;[x=r+1/3,3*r/d-x,1/d-1]

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$x^3+y^3+z^3=R$. Proceedings of the Edinburgh Mathematical Society, 2(2), 92-100.

$$r=\left(\frac{27r^3+1}{27r^2-9r+3}\right)^3+\left(\frac{-27r^3+9r-1}{27r^2-9r+3}\right)^3+\left(\frac{-27r^2+9r}{27r^2-9r+3}\right)^3$$

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Haskell, 95 89 76 69 68 bytes

• -18 bytes thanks to H.PWiz
• -1 byte thanks to Christian Sievers

f x=[w|n<-[1..],w<-mapM(\_->[-n,1/n-n..n])"IOU",x==sum((^3)<$>w)]!!0

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Simple bruteforce solution. It tests all triples of rational numbers of the form

$$\left(\frac{a_1}{n},\frac{a_2}{n},\frac{a_3}{n}\right)\qquad\text{with }-n\le\frac{a_i}{n}\le n.$$

• We can always assume that the three rational numbers have the same denominator, since $$\left(\frac{a_1}{n_1},\frac{a_2}{n_2},\frac{a_3}{n_3}\right)=\left(\frac{a_1n_2n_3}{n_1n_2n_3},\frac{a_2n_1n_3}{n_1n_2n_3},\frac{a_3n_1n_2}{n_1n_2n_3}\right).$$
• We can always assume that $-n\le\frac{a_i}{n}\le n$, since $$\frac{a_i}{n}=\frac{a_iN}{nN}$$ for any arbitrarily large integer $N$.

Husk, 14 bytes

ḟo=⁰ṁ^3π3×/NİZ

Simple brute force solution. Try it online!

Explanation

Division in Husk uses rational numbers by default and Cartesian products work correctly for infinite lists, making this a very straightforward program.

ḟo=⁰ṁ^3π3×/NİZ
            İZ  Integers: [0,1,-1,2,-2,3,-3...
           N    Natural numbers: [1,2,3,4,5...
         ×/     Mix by division: [0,1,0,-1,1/2,0,2,-1/2,1/3...
                This list contains n/m for every integer n and natural m.
       π3       All triples: [[0,0,0],[0,0,1],[1,0,0]...
ḟ               Find the first one
    ṁ^3         whose sum of cubes
 o=⁰            equals the input.

JavaScript (Node.js), 73 bytes

Takes input as (p)(q), where $p$ and $q$ are BigInt literals.

Returns [[p1,q1],[p2,q2],[p3,q3]] such that $\frac{p}{q}=\left(\frac{p_1}{q_1}\right)^3+\left(\frac{p_2}{q_2}\right)^3+\left(\frac{p_3}{q_3}\right)^3$.

p=>q=>[x=p*(y=p*(p*=9n*q*q)*3n/q)/q+(q*=q*q),p-x,p-=y].map(x=>[x,3n*q-p])

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Derived from H. W. Richmond (1930), On Rational Solutions of x³ + y³ +z³ = R.

Haskell, 70 bytes

In An introduction to the Theory of Numbers (by Hardy and Wright) there is an construction that even includes a rational parameter. For golfing purposes I just set this parameter to 1, and tried reducing as much as possible. This results in the formula

$$r \mapsto \left[ {{r^3-648\,r^2+77760\,r+373248}\over{72\,\left(r+72\right)^2 }} , {{12\,\left(r-72\right)\,r}\over{\left(r+72\right)^2}} , -{{r^2 -720\,r+5184}\over{72\,\left(r+72\right)}} \right]$$

f r|t<-r/72,c<-t+1,v<-24*t/c^3,a<-(v*t-1)*c=((a+v*c+c)/2-)<$>[a,v*c,c]

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perl -Mbigrat -nE, 85 bytes

$_=eval;($a,$b)=($_*9,$_**2*27);$c=$b*$_;say for map$_/($b-$a+3),$c+1,-$c+$a-1,-$b+$a

You can save 8 bytes (the leading $_=eval;) if you know the input is an integer; this part is needed to have the program grok an input of the form 308/1728. Input is read from STDIN. I'm using the formula given by @alephalpha.