\(A,\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3=x^3-y^3+2y^3=x^3+y^3\)
\(B,\frac{2x}{3x-3y}:\frac{x^2}{x-y}=\frac{2x}{3.\left(x-y\right)}.\frac{x-y}{x^2}=\frac{2x\left(x-y\right)}{3x^2\left(x-y\right)}=\frac{2}{3x}\)
\(C,\left(x-2\right)\left(\frac{3}{x+2}-\frac{5}{2x-4}+\frac{8}{x^2-4}\right)=\left(x-2\right)\left(\frac{3}{x+2}-\frac{5}{2.\left(x-2\right)}+\frac{8}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(x-2\right).\frac{6.\left(x-2\right)-5.\left(x+2\right)+8.2}{2.\left(x+2\right)\left(x-2\right)}=\left(x-2\right).\frac{6x-12-5x-10+16}{2.\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x-6\right)}{2.\left(x+2\right)\left(x-2\right)}=\frac{x-6}{2.\left(x+2\right)}\)
\(D,\left(x-3\right)\left(\frac{2}{x+3}-\frac{3}{2x-6}+\frac{9}{x^2-9}\right)=\left(x-3\right)\left(\frac{2}{x+3}-\frac{3}{2.\left(x-3\right)}+\frac{9}{\left(x+3\right)\left(x-3\right)}\right)\)
\(=\left(x-3\right).\frac{4\left(x-3\right)-3\left(x+3\right)+9.2}{2.\left(x+3\right)\left(x-3\right)}=\left(x-3\right).\frac{4x-12-3x-9+18}{2\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(x-3\right)}{2.\left(x+3\right)\left(x-3\right)}=\frac{x-3}{2.\left(x+3\right)}\)
