问题标题：
x7+1因式分解,
问题描述：
x7+1因式分解,
马启波回答：
x^7+1
首先增加一次项
=x^7-x+x+1
=x(x^6-1)+(x+1)
接着做平方差、立方和、立方差分解
=x(x'"-1)(x"'+1)+(x+1)
=x(x-1)(x"+x+1)(x+1)(x"-x+1)+(x+1)
提取公因式(x+1)
=(x+1)[x(x-1)(x"+1+x)(x"+1-x)+1]
=(x+1)[x(x-1)(x^4+2x"+1-x")+1]
再用平方差相乘
=(x+1)[(x"-x)(x^4+x"+1)+1]
=(x+1)[x^6+x^4+x"-x^5-x^3-x+1]
=(x+1)(x^6-x^5+x^4-x"'+x"-x+1)
或者增加三次项
=x^7-x"'+x"'+1
=x"'(x^4-1)+(x+1)(x"-x+1)
=x"'(x"-1)(x"+1)+(x+1)(x"-x+1)
=x"'(x-1)(x+1)(x"+1)+(x+1)(x"-x+1)
=(x+1)[x"'(x-1)(x"+1)+(x"-x+1)]
=(x+1)[x"'(x"'-x"+x-1)+x"-x+1]
=(x+1)(x^6-x^5+x^4-x^3+x"-x+1)
增加四次项也行
=x^7+x^4-x^4+1
=x^4(x"'+1)-(x^4-1)
=x^4(x+1)(x"-x+1)-(x"-1)(x"+1)
=(x+1)(x^6-x^5+x^4)-(x+1)(x-1)(x"+1)
=(x+1)[(x^6-x^5+x^4)-(x"+1)(x-1)]
=(x+1)[(x^6-x^5+x^4)-(x"'-x"+x-1)]
=(x+1)(x^6-x^5+x^4-x"'+x"-x+1)
还可以增加六次项
=x^7+x^6-x^6+1
=x^6(x+1)-(x^6-1)
=x^6(x+1)-(x"'+1)(x"'-1)
=x^6(x+1)-(x+1)(x"+1-x)(x"+1+x)(x-1)
=x^6(x+1)-(x+1)(x^4+2x"+1-x")(x-1)
=(x+1)[x^6-(x^4+x"+1)(x-1)]
=(x+1)[x^6-(x^5+x^3+x-x^4-x"-1)]
=(x+1)[x^6-x^5-x^3-x+x^4+x"+1]
=(x+1)(x^6-x^5+x^4-x^3+x"-x+1)
总之,7=6+1=2X3+1
还有,7=4+3=2X2+3
利用一次项、三次项、四次项、六次项做变化
都可以使用平方差、立方和、立方差,进行分组分解
可是二次项、五次项就不行了
这样看来,还是利用四次项做变化最方便
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