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Possible Duplicate:
Intuitive explanation for the identity $\sum\limits_{k=1}^n {k^3} = \left(\sum\limits_{k=1}^n k\right)^2$

How can one prove that $$(1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$$

Let $T_1=(1+2+3+\cdots+n)=\dfrac{n(n+1)}{2}$.

Now, $T_2=(1^2+2^2+3^2+\cdots+n^2)$.

$T_2=((1^2+n^2)+(2^2+(n-1)^2)+(3^2+(n-2)^2)+\cdots+(\frac{n+1}{2})^2)$, when $n$ is odd

$T_2=((1^2+n^2)+(2^2+(n-1)^2)+(3^2+(n-2)^2)+\cdots+(k^2+(n-k+1)^2))$, when $n$ is even, where, $k=\frac{n}{2}$

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HOLYBIBLETHE
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    @Sidd: I know this one by heart because I was the first to ask it a year and a half ago. It has been asked many times, and at some point there were 3 or 4 existing duplicates. I merged all of them into the single most upvoted question. (Closing my original in the process) I will merge this one in as well in a few days to preserve Jennifer Dylan's answer. – Eric Naslund Aug 29 '12 at 23:10
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    @EricNaslund I respectfully disagree with your duplicate suggestion. I think this question is the actual duplicate, since it contains an actual inductive proof (e.g. alok's answer) rather than intuitive explanations. –  Aug 29 '12 at 23:56

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