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Is there an intuitive way to get cubic sum? From this post: combination of quadratic and cubic series and Wikipedia: Faulhaber formula, I get $$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$ I think the cubic sum is squaring the arithmetic sum $$1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$$ But how to prove it? Please help me. Grazie!

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Maybe this will help you visualize it:

Source.

or this one which is clearer:

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  • $\begingroup$ Very nice! Thanks for your answer :) $\endgroup$ – L Lawliet Jul 24 '14 at 14:01
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    $\begingroup$ @LLawliet You're welcome, glad I could help! ;) $\endgroup$ – Hakim Jul 24 '14 at 14:01
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We have $$ \sum_{k=1}^{n}k^3 = 1 + 8 + 27 + \ldots + n^3 = \\ \underbrace{1}_{1^3} + \underbrace{3+5}_{2^3} + \underbrace{7 + 9 + 11}_{3^3} + \underbrace{13 + 15 + 17 + 19}_{4^3} + \ldots = \\ \underbrace{\underbrace{\underbrace{1}_{1^2} + 3}_{2^2} + 5}_{3^2} + \ldots $$ which is $$ \left( \sum_{k=1}^{n}k \right)^2 $$

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    $\begingroup$ This is also very nice! Thanks! $\endgroup$ – L Lawliet Jul 24 '14 at 14:05
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    $\begingroup$ Yes, I find it quite amusing, just messing around with numbers and see what comes out =) $\endgroup$ – Noxet Jul 25 '14 at 8:34

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