请教做法！谢谢！.

设x-y=a、y-z=b、z-x=c,则a+b+c=0 ==> (a+b)^5=-c^5,
将其展开并移项得
a^5+b^5+c^5
=5a^4b+10a^3b^2+10a^2b^3+5ab^4
=5ab(a+b)(a^2+ab+b^2)
=5abc(c^2-ab),
以所设代入，
即：(x-y)^5+(y-z)^5+(z-x)^5=5(x-y)(y-z)(z-x)[(x^2+y^2+z^2-xy-yz-zx)。.

(a+b)^5
=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5.

我们这几天在学而思也在学这个，老师教了个杨辉三角，很好用，你可以网上查一下.

谢谢！.