Từ \(x+y=1\Rightarrow x=1-y\)
\(\Rightarrow y=1-x\)
Biến đổi \(\frac{y}{x^3-1}=\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(1-x\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{1}{x^2+x+1}\)
\(\frac{x}{y^3-1}=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}=\frac{\left(1-y\right)}{\left(y-1\right)\left(y^2+y+1\right)}=-\frac{\left(y-1\right)}{\left(y-1\right)\left(y^2+y+1\right)}=-\frac{1}{y^2+y+1}\)
Ta có \(\frac{y}{x^3-1}-\frac{x}{y^3-1}=\frac{-1}{x^2+x+1}-\frac{-1}{y^2+y+1}=\frac{-y^2-y-1+x^2+x+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\)
\(=\frac{\left(x^2-y^2\right)+\left(x-y\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=\frac{\left(x-y\right)\left(x+y\right)+\left(x-y\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=\frac{\left(x-y\right)\left(x+y+1\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\)
\(=\frac{\left(x-y\right)\left(1+1\right)}{x^2y^2+xy^2+y^2+xy+y+x^2+x+1}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+x+y+1\right)+\left(x^2+y^2\right)+\left(x^2y+xy^2+xy\right)}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+2\right)+\left(x^2+y^2\right)+xy\left(x+y+1\right)}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+2\right)+\left(x^2+y^2\right)+xy\left(1+1\right)}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+2\right)+x^2+y^2+2xy}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+2\right)+\left(x+y\right)^2}\)
\(=\frac{2\left(x-y\right)}{x^2y^2+2+1}\)
\(=\frac{2\left(x-y\right)}{x^2y^2+3}\)
Vậy \(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{2\left(x-y\right)}{x^2y^2+3}\) với \(x+y=1\&xy\ne0\)
