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Let's provide a simple numerical example to get an idea of what is involved. We start with the general equation $a^3 +b^3 +c^3=d^3$. We then take two known particular solutions and add them in the way mentioned below to create totally different equations in the number of cubes involved in the left hand side and the right hand side.

$3^3 + 4^3 + 5^3 = 6^3$
$2^3 + 12^3 + 16^3 = 18^3$

The resulting equation by adding the above two equations term by term is ( adding the same left hand sides and adding the same right hand sides to get the following equation):

$3^3 + 4^3 + 5^3 + 2^3 + 12^3 + 16^3 =6^3 + 18^3$

and/or if we change the order to add the two equations in the following way. Here we add a left hand side of the very first equation to the right hand side of the second equation)

$3^3 +4^3 + 5^3 =6^3$ to add term by term to:
$18^3 = 2^3 +12^3 + 16^3$ to get:

$3^3 + 4^3 + 5^3 + 18^3 = 6^3 + 2^3 + 12^3 + 16^3$

I would like to know if any known method used to solve the first two equations can be used to solve the resulting equation(s)? I suspect that in some cases, it will be possible to re-create the two original equations then use any known method to solve them and consequently to solve the last two equations.

Fo reference, I provide the following links:

Sums of three cubes of form $a^3+b^3+c^3=(c+1)^3$?

Diophantine Equation $a^3+b^3+c^3=d^3$

user25406
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  • You've got a $2^3$ when you want $3^3$ in the first result of adding. – Thomas Andrews Oct 23 '23 at 16:23
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    The first two equations? Usually, equations have variables, and the two solution sets you've given seem solutions to the same one equation. – Thomas Andrews Oct 23 '23 at 16:25
  • Sorry, I will correct it. – user25406 Oct 23 '23 at 17:34
  • You've fixed the numeric typo, but you still haven't told us what equations you are talking about. We can only guess. – Thomas Andrews Oct 23 '23 at 17:37
  • @ThomasAndrews, can you write down the equation you mean by "the same one equation"? – user25406 Oct 23 '23 at 17:37
  • I don't have the feeling that the two equations resulting from the addition of the very first two equations are the same since the third equation is two cubes as a sum of six cubes but the last equation is a sum of four cubes as a sum of four different cubes involving of course the same cubes. – user25406 Oct 23 '23 at 17:44
  • If you don't know what equations you are talking about, how are we to help you? The one equation I thought you were talking about, when you first mentioned the "two equations" is $a^3+b^3+c^3=d^3.$ – Thomas Andrews Oct 23 '23 at 17:58
  • yes, and I am just adding two different numerical values of the equation you wrote down to create two new equations that are a sum of two cubes as a sum of 6 cubes and a sum of for cubes as a sum of four cubes. The numerical vales of $a,b,c,d$ do not change but the final two equations are very different from the starting equations. – user25406 Oct 23 '23 at 18:02

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I think you're asking about the Diophantine equation $$ a^3 + b^3 + c^3 + d^3 = e^3 + f^3 + g^3 + h^3$$ As you've noticed, this has solutions where $a^3 + b^3 + c^3 = e^3$ and $d^3 = f^3 + g^3 + h^3$. However, it has lots of other solutions that are not of this type. For example, $$ 2^3 + 3^3 + 4^3 + 19^3 = 1^3 +5^3 + 10^3 + 18^3 $$ but no three of the terms on the left add to a term on the right.

Robert Israel
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  • So the solution I provided for a sum of four cubes adding up to a sum of four different cubes when starting from two sums of 3 cubes adding up to a cube $3^3+4^3+5^3 +18^3= 6^3 + 2^3 + 12^3 +16^3$ is not a general solution. I do not know if there is a method to solve the case you provided when "no three of the terms on the left add to a term on the right". Thanks for pointing that out with a numerical example. – user25406 Oct 23 '23 at 19:55