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I was watching a Numberphile video and I saw how Mathematicians were able to express almost any number as the sum of three cubes$$\begin{align*}1 & =1^3+0^3+0^3\\ & =9^3+10^3+(-12)^3\\29 & =3^3+1^3+1^3\\53 & =27^3+27^3+(-1)^3\\51 & =659^3+602^3+(-796)^3\end{align*}$$However, I'm wondering how do you generate these types of values. It's obvious for small-valued numbers, but some can get really exhausting$$30=(2\,220\,433\,932)^3+(-2\,218\,888\,517)^3+(-283\,069\,966)^3$$

Question: How are these kinds of solutions generated?

Frank
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    https://mathoverflow.net/questions/138886/which-integers-can-be-expressed-as-a-sum-of-three-cubes-in-infinitely-many-ways – Moo Aug 19 '17 at 21:57
  • It's hard to express a number as a sum of cubes but it's easy to start from a any number and find the sum of three cubes of a larger number. Example: $8^3=512= (3+5)^3=3^3 +33^25 + 335^2 + 5^3=3^3 + 5^3 + 360$, since $360>7^3=343$ so we pick $8^3=512$ so we need to add $(512-360)=152$ to both sides of the original equation to get a cube. $8^3 +152=664 =3^3 +5^3 +8^3$. We started with $8^3$ and turned it into a number which is exactly the sum of $3$ cubes. But it's not easier to start with a general number and find its sum of 3 cubes. – user25406 Dec 15 '22 at 21:37
  • By the way, if we wanted to get the following sum of $3$ cubes equal to $3^3 + 5^3 +17^3$, then we would have to add $17^3-360=4553$ to both sides so that we end up with $8^3 +4553 = 5065 = 3^3 + 5^3 + 17^3$. We could also have started with $8^3=(1+7)^3=(2+6)^3$... – user25406 Dec 15 '22 at 22:19

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