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Prove that for every integer $ n \geq 1$, we have

$\displaystyle \sum_{j=1}^n j^3 = \left(\dfrac{n(n+1)}{2}\right)^2$

I know how to prove an induction proof, but I just can't get the algebra down on this problem. Can anyone help?

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$$\left(\frac{n(n+1)}2\right)^2+(n+1)^3$$ $$=\frac{n^2(n+1)^2}4+(n+1)^3$$ $$=\frac{n^2(n+1)^2}4+\frac{4(n+1)^3}4$$

$$=\frac{n^2(n+1)^2}4+\frac{4(n+1)(n+1)^2}4$$$$=\frac{(n+1)^2}4\left[n^2+4(n+1)\right]$$ $$=\frac{(n+1)^2}4(n^2+4n+4)$$ $$=\frac{(n+1)^2}4(n+2)^2$$ $$=\frac{(n+1)^2(n+2)^2}4$$ $$=\left(\frac{(n+1)(n+2)}2\right)^2.$$

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