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How to find this summation $\sum ^{999}_{n=1}\dfrac {1}{\sqrt [3] {n^{2}+2n+1}+\sqrt [3] {n^{2}+n}+\sqrt [3] {n^{2}}}$

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Notice that $a^3-b^3= (a-b)(a^2+ab+b^2)$.

This gives $$ \dfrac {1}{\sqrt [3] {n^{2}+2n+1}+\sqrt [3] {n^{2}+n}+\sqrt [3] {n^{2}}}=\frac{\sqrt [3]{n+1}-\sqrt [3]n}{n+1 -n} $$

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Is it telescopic?? The answer is 9 right? – user109004 Dec 11 '13 at 10:12
Yes. That's right. – i707107 Dec 11 '13 at 10:13
Thanks for your hint ^^ – user109004 Dec 11 '13 at 10:19

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