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By using $x^3-(x-y)^3 = y(3x^2-3xy+y^2)$, find the value of $$(1^3-2^3+3^3-\cdots+9^3)\times3 + (10^3-11^3+12^3-\cdots-19^3).$$

I find different solutions regarding the alternate sum of cubes here, but none of the solutions shows a clear usage of the identity. I was thinking about grouping $10^3$ and $11^3$ first, but that would give me $x=10$ and $y=-1$ and it does not seem to be useful. I would appreciate any help.

  • I am looking for solutions without using the sum of cubes identity. Any hint is welcome.
Nighty
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1 Answers1

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Hint:

You have a formula for the sum of cubes, which I suppose you know. For the alternate sum of cubes, let me take the first alternate sum as an example: \begin{align} 1^3-2^3+3^3-\cdots+9^3&=(1^3+2^3+3^3+\cdots+9^3)-2(2^3+4^3+6^3+8^3)\\ &=(1^3+2^3+3^3+\cdots+9^3)-2\cdot 2^3(1^3+2^3+3^3+4^3) \end{align} Can you continue?

Bernard
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  • Thanks for the reply. I understand this method, but I am curious about the solution using the given identity. In other words, one should be able to do it without the sum of cubes formula. – Nighty Sep 22 '19 at 15:21
  • Oh! Sorry, I had overlooked the question. I'll think about this other method. – Bernard Sep 22 '19 at 15:23