2楼saicjane
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发表于 2014-1-15 13:25
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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)。.