First note that $1-x_i=1-{i\over 101}={{101-i}\over 101}=x_{101-i}$ Now \begin{eqnarray*}\sum_{i=0}^{101}\frac{x_i^3}{1-3x_i+3x_i^2} &=& \sum_{i=0}^{101}\frac{x_i^3}{(1-x_i)^3+x_i^3}\\ &=& \sum_{i=0}^{101}\frac{x_i^3}{x_{101-i}^3+x_i^3}\\ &=& \sum_{i=0}^{50}\left(\frac{x_i^3}{x_{101-i}^3+x_i^3}+\frac{x_{101-i}^3}{x_{101-i}^3+x_i^3}\right)\\ &=& \sum_{i=0}^{50} 1\\ &=& 51\end{eqnarray*} Generally, $\sum_{i=0}^{2n-1}\frac{x_i^3}{1-3x_i+3x_i^2}=n$ where $x_i=\frac{i}{2n-1}$

More generally (for 2n) the sum gonna be equal to n+1 ????

The ith term is $(x_i)^∧3/(1 - 3x_i + 3(x_i)^2) = (x_i)^3/( (1 -(x_i))^3 + (x_i)^3) = i^3/( (101 - i)^3 + i^3)$ Hence the total sum is $51$.

@above the $\LaTeX$ is incorrect. try using ^ for an exponent and {} after the exponent. FOr example, if you wanted $i^{7i \cdot (I + 1)}$, you would do $i^{7i \cdot (I + 1)}$

Let $$a_k=\frac{k^3}{101^3-3\cdot 101^2k+3\cdot 101k^2}.$$ Claim: $$a_k+a_{101-k}=1.$$Note that the denominator of $101^3-3\cdot 101^2k+3\cdot 101k^2$ is symmetric around $$k=\frac{3\cdot101^2}{2\cdot 3\cdot 101}=101/2,$$so $a_k$ and $a_{101-k}$ have the same denominator. Furthermore, $k^3+(101-k)^3$ is equal to that common denominator, so their sum is 1. Thus, the answer is 51. remark: this is motivated by trying small numbers in place of $101$, you will see that the terms in the sum come in pairs that add to 1 and this easily generalizes

Nice! My first APMO problem. We see $\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{x_i^3+(1-x_i)^3}$; in particular, we are motivated to see that $1-x_i=x_{101-i}$. Combining, we get $$LHS=\sum_{i=0}^{50}\frac{x_i^3}{x_i^3+x_{101-i}^3}+\frac_{x_{101-i}^3}{x_i^3+x_{101-i}^3}=51.\blacksquare$$ im not sure whats wrong with my latex can someone fix

@above huashiliao2020 wrote: Nice! My first APMO problem. We see $\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{x_i^3+(1-x_i)^3}$; in particular, we are motivated to see that $1-x_i=x_{101-i}$. Combining, we get $$\text{LHS} =\sum_{i=0}^{50}\frac{x_i^3}{x_i^3+x_{101-i}^3}+\frac{x_{101-i}^3}{x_i^3+x_{101-i}^3}=51.\blacksquare$$ im not sure whats wrong with my latex can someone fix