Question

初一数学合并同类项题

2(x-3y)↑2 - 3(x-3y)↑3 - (3y-x)↑2 + (3y-x)↑3

Answer 1

2(x-3y)↑2 - 3(x-3y)↑3 - (3y-x)↑2 + (3y-x)↑3
=[2(x-3y)↑2- (3y-x)↑2]+[- 3(x-3y)↑3 +(3y-x)↑3](根据加法的交换律和结合律)
=[2(x-3y)↑2- (x-3y)↑2]+[3(3y-x)↑3 +(3y-x)↑3]
=(x-3y)↑2+4(3y-x)↑3或者=（x-3y)↑2-4(x-3y)↑3
合并同类项法则：把各项系数相加，字母和字母的指数不变。

Answer 2

楼上的错了
2(x-3y)↑2 - 3(x-3y)↑3 - (3y-x)↑2 + (3y-x)↑3
=2(x-3y)↑2- (3y-x)↑2- 3(x-3y)↑3 + (3y-x)↑3
=2(x-3y)↑2- (x-3y)↑2- 3(x-3y)↑3 -(x-3y)↑3
=(x-3y)↑2- 4(x-3y)↑3

Answer 3

2(x-3y)↑2 - 3(x-3y)↑3 - (3y-x)↑2 + (3y-x)↑3
=2(x-3y)↑2- (3y-x)↑2- 3(x-3y)↑3 + (3y-x)↑3
=(x-3y)↑2- 4(x-3y)↑3
=(x-3y)↑2(1-4x+12y)

Answer 4

=2(x-3y)↑2 - 3(x-3y)↑3-(x-3y)↑2-(x-3y)↑3
=(x-3y)↑2-4(x-3y)↑3