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.
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?