He is your Opposite, your Negative.

First notice, $a_{101-n} = \frac{101-n}{101} = 1 - \frac{n}{101} = 1 - a_n$

$\Rightarrow a_n = 1 -a_{101-n}$

$$

$\frac{a_n^3}{1-3a_n+3a_n^2} + \frac{a_{101-n}^3}{1-3a_{101-n}+3a_{101-n}^2}$

$= \frac{a_n^3}{(1-a_n)^3+a_n^3} + \frac{a_{101-n}^3}{(1-a_{101-n})^3+a_{101-n}^3}$

$= \frac{a_n^3}{(1-a_n)^3+a_n^3} + \frac{(1 - a_n)^3}{a_n^3+(1-a_n)^3}$

$= 1$

Here, the summation of $n$ th and $(101-n)$ th terms are $1$ . Thus the sum from $n=1$ to $100$ is $50$ .

The $101$ th term is $\frac{a_{101}}{1-3a_{101}+3a_{101}^2} = \frac{1}{1-3+3} = 1$ .

So, total sum = $\boxed{51}$