(x+y)(x⁴-x³y+x²y²-xy³+y⁴)
解析：
//因式分解
x^5+y^5
显然，y=-x时，x^5+y^5=0
故，分解结果中含“x+y”
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x^5+y^5
=(x^5+x⁴y)+(y^5-x⁴y)
=x⁴(x+y)+y(y⁴-x⁴)
=x⁴(x+y)-y(x²+y²)(x-y)(x+y)
=(x+y)[x⁴-y(x-y)(x²+y²)]
=(x+y)(x⁴-x³y+x²y²-xy³+y⁴)
x^5+y^5=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
x^5-y^5
=(x-y)(x^4+x??y+x??y??+xy??+y^4)
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