Question

Помогите решить,задание 713(б)

Answer 1

$3x^3-2y^3-6x^2y^2+xy=(3x^3-6x^2y^2)+(xy-2y^3)=3x^2*(x-2y^2)+y*(x-2y^2)=(x-2y^2)*(3x^2+y)=(\frac{2}{3} -2*(\frac{1}{2})^2)*(3*(\frac{2}{3})^2+\frac{1}{2})=(\frac{2}{3} -\frac{1}{2})*(\frac{4}{3}+\frac{1}{2})=(\frac{4}{6} -\frac{3}{6})*(\frac{8}{6}+\frac{3}{6})=\frac{1}{6}*\frac{11}{6}=\frac{11}{36}$