Решите ^-cтепень а)(b^2+1)(3b−b^3)−(3b−b^5)= б)(2x−3y)(3x+2y)−(6x^2−5xy)=
A) (b^2+1)(3b-b^3)-(3b-b^5) (3b^3-b^5+3b-b^3)-(3b-b^5) -b^5+3b-3b+b^5 Ответ: 0 b) (2x-3y)(3x+2y)-(6x^2-5xy) (6x^2+4xy-6xy-6y^2)-(6x^2-5xy) 6x^2-2xy-6y^2-6x^2+5xy 3xy-6y^2 3y(x-2y) Ответ; 3y(x-2y)