Question

（y-x)3次幂（x-y)n次幂+(x-y）n+1次幂（y-x）2次幂,

Answer 1

[(y-x)^3][(x-y)^n]+[(x-y)^(n+1)][(y-x)^2]

=-(x-y)^(n+3)+(x-y)^(n+3）=0