Question 1119603
a) Since {{{(a-b)^2=(b-a)^2}}}
{{{(a-b)^2(2a-3b-3a+5b)+(a+b)^2(a-2b)=(a-b)^2(-a+2b)+(a+b)^2(a-2b)=-(a-2b)(-(a-b)^2+(a+b)^2)=-(a-2b)(-a^2+2ab-b^2+a^2+2ab+b^2)=-(a-2b)(-4ab)=4ab(a-2b)}}}

b){{{(xy+1)^2-(x+y)^2=x^2y^2+2xy+1-x^2-2xy-y^2=x^2y^2+1-x^2-y^2=(x^2y^2-x^2)+(1-y^2)=x^2(y^2-1)-(y^2-1)=(y^2-1)(x^2-1)=(y-1)(y+1)(x-1)(x+1)}}}