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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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    @EricNaslund I respectfully disagree with your duplicate suggestion. I think [this question](http://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct?lq=1) is the actual duplicate, since it contains an actual inductive proof (e.g. alok's answer) rather than intuitive *explanations*.2012-08-29

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