Question

因式分解奥数题

Answer 1

整式 (a+b+c)^4 - (b+c)^4 - (c+a)^4 - (a+b)^4 + a^4 + b^4 + c^4 分解因式，
结果是 12abc( a + b + c )，
这个结果究竟是怎么得到的呢？
既然都是四次方，就还是想想平方差，分组分解吧，
= (a+b+c)^4 - (a + b)^4 + c^4 - (b + c)^4 + a^4 - (c + a)^4 + b^4
= (a+b+c)^4 - [ (a + b)^4 - c^4 ] - [ (b + c)^4 - a^4 ] - [ (c + a)^4 - b^4 ]
这个式子太长，我们就先分别处理后面三组

第一组 - [ (a + b)^4 - c^4 ]
= - [ (a + b)" + c" ][ (a + b) - c ][ (a + b) + c ]
= - [ (a" +b" +c") + 2ab ][ (a + b + c) - 2c ]( a + b + c )
= - ( a + b + c )[ (a" +b" +c")(a + b + c) + 2ab(a + b + c) - 2c(a" +b" +c") - 4abc ]

第二组 - [ (b + c)^4 - a^4 ]
= - [ (b + c)" + a" ][ (b + c) - a ][ (b + c) + a ]
= - [ (a" +b" +c") + 2bc ][ (a + b + c) - 2a ]( a + b + c )
= - ( a + b + c )[ (a" +b" +c")(a + b + c) + 2bc(a + b + c) - 2a(a" +b" +c") - 4abc ]

第三组 - [ (c + a)^4 - b^4 ]
= - [ (a + c)" + b" ][ (a + c) - b ][ (a + c) + b ]
= - [ (a" +b" +c") + 2ac ][ (a + b + c) - 2b ]( a + b + c )
= - ( a + b + c )[ (a" +b" +c")(a + b + c) + 2ac(a + b + c) - 2b(a" +b" +c") - 4abc ]

这样，就找到一个公因式 -(a + b + c)，
这三组各自的另一个因式里面，除了一头一尾都是
(a" +b" +c")(a + b + c) 和 - 4abc，
它们 “合并同类项” 就是
3(a" +b" +c")(a + b + c) - 12abc

其余的式子 “提公因式”，“合并同类项” 就是
( 2ab + 2bc + 2ac )(a + b + c) - ( 2a + 2b + 2c )(a" +b" +c")
= ( 2ab + 2bc + 2ac )(a + b + c) - 2(a + b + c)(a" +b" +c")
合并前面的
3(a" +b" +c")(a + b + c) - 12abc，
后面这三组就是
= - (a + b + c)[ (a" +b" +c")(a + b + c) + ( 2ab + 2bc + 2ac )(a + b + c) - 12abc ]
= - (a + b + c)[ (a" +b" +c" + 2bc + 2ab + 2ac )(a + b + c) - 12abc ]
= - (a + b + c){ [ a" + (b + c)" + 2a(b + c) ](a + b + c) - 12abc }
= - [ (a + b + c)"' - 12abc ](a + b + c)

整个式子，就是
= (a + b + c)^4 - [ (a + b + c)"' - 12abc ](a + b + c)
= [ (a + b + c)"' - (a + b + c)"' + 12abc ](a + b + c)
= 12abc(a + b + c)