S. Ryley proved following theorem in 1825:

Every rational number can be expressed as a sum of three rational cubes.

Challenge

Given some rational number $r in mathbb Q $ find three rational numbers $a,b,c in mathbb Q$ such that $$r= a^3+b^3+c^3.$$

Details

Your submission should be able to compute a solution for every input given enough time and memory, that means having for instance two 32-bit int representing a fraction is not sufficient.

Examples

$$ begin{align}
30 &= 3982933876681^3 - 636600549515^3 - 3977505554546^3 \
52 &= 60702901317^3 + 23961292454^3 - 61922712865^3 \
frac{307}{1728} &= left(frac12right)^3 + left(frac13right)^3 + left(frac14right)^3 \
0 &= 0^3 + 0^3 + 0^3 \
1 &= left(frac12right)^3 + left(frac23right)^3 + left(frac56right)^3\
42 &= left(frac{1810423}{509232}right)^3 + left(frac{-14952}{10609}right)^3 + left(frac{-2545}{4944}right)^3
end{align}$$

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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The same length, the same formula:

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

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This formula is given in:
Richmond, H. (1930). On Rational Solutions of $x^3+y^3+z^3=R$. Proceedings of the Edinburgh Mathematical Society, 2(2), 92-100.

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

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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
$$
left(frac{a_1}{n},frac{a_2}{n},frac{a_3}{n}right)qquadtext{with }-nlefrac{a_i}{n}le n.
$$

• 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 $-nlefrac{a_i}{n}le n$, 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.
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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

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

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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