$f(x) = \frac{x^3}{1 -3x +3x^2}\Rightarrow f(1-x) = \frac{(1-x)^3}{1 -3(1-x) +3(1-x)^2}= \frac{1 -3x +3x^2-x^3}{1 -3x +3x^2}=1- \frac{x^3}{1 -3x +3x^2}=1-f(x).$
Suy ra $f(1-x)+f(x)=1$.
Ta có
$A = \left[ {f(\frac{1}{2014})+ f(\frac{2013}{2014})} \right]+ \left[ {f(\frac{2}{2014})+ f(\frac{2012}{2014})} \right]+\ldots+ \left[ {f(\frac{1006}{2014})+ f(\frac{1008}{2014})} \right]+ f(\frac{1007}{2014})$
$A = 1+1+\ldots+1+ f(\frac{1}{2}) $
$A = 1006+ \frac{1}{2}=1006\frac12.$.