Ta có:
\(A=x\left(x^3-1\right)-y\left(y^3-1\right)=x^4-x-y^4+y\)
\(=\left(x^4-y^4\right)+\left(-x+y\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+y^2-1\right)=\left(x-y\right)\left[\left(x+y\right)^2-2xy-1\right]\)
\(=-2xy\left(x-y\right)\)
\(B=\left(y^3-1\right)\left(x^3-1\right)=x^3y^3-x^3-y^3+1\)
\(=x^3y^3+1-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3y^3+1-\left[\left(x+y\right)^2-3xy\right]\)
\(=xy\left(x^2y^2+3\right)\)
Từ đó ta có:
\(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\dfrac{x\left(x^3-1\right)-y\left(y^3-1\right)}{\left(y^3-1\right)\left(x^3-1\right)}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\dfrac{-2xy\left(x-y\right)}{xy\left(x^2y^2+3\right)}+\dfrac{2\left(x-y\right)}{x^2y^2+3}=-\dfrac{2\left(x-y\right)}{x^2y^2+3}+\dfrac{2\left(x-y\right)}{x^2y^2+3}=0\)