<s>[INST] Find $f(1)+f(3)+f(5)+\dots+ f(999)$ where $f$ is a given function 
Given $$f(x)= \frac {1}
{\sqrt[3] {x^2+2x+1} + \sqrt[3] {x^2-1} + \sqrt[3] {x^2-2x+1}}$$
  and 
  $$E= f(1)+f(3)+f(5)+\dots+ f(999).$$ Then find the value of $E$.

My work :- 
Let $\sqrt[3] {x+1}= a$, $\sqrt[3] {x-1}=b $ 
Then the equation reduces to 
$$f(x)= \frac {1}{a^2+b^2+ab}.$$ 
Now for this expression I tried breaking into partial fractions to that the sum telescopes to some finite quantity but the cube root terms restrict this straightforward way. Thanks in advance for the help. Also if someone finds a way to rationalize the denominator, please share. 
 [/INST] Hint. Since $(a^3-b^3)=(a-b)(a^2+ab+b^2)$, it follows that
$$f(x)=\frac{1}{a^2+ab+b^2}=\frac{a-b}{a^3-b^3}=\frac{\sqrt[3] {x+1}-\sqrt[3] {x-1}}{2}.$$
Therefore the given sum is telescopic:
$$\sum_{k=0}^{n-1}f(2k+1)=\frac{1}{2}\left(\sum_{k=0}^{n-1}\sqrt[3] {2(k+1)}-\sum_{k=0}^{n-1}\sqrt[3] {2k}\right)=\frac{\sqrt[3] {2n}}{2}.$$
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