問題 14 次の式を因数分解せよ。
(1) (a+b)(b+c)(c+a) + abc
(2) a(b-c)²+b(c-a)² + c(a - b)²+8abc
(1) (a+b)(b+c)(c+a)+ abc
=
==
=
(b+c){(a+b)(a+c)}+(bc)a
(b+c){a²+(b+c)a+bc}+ (bc)a
(b+c)a²+{(b+c)2+bc}a+bc(b+c)
= {(b+c)a+bc}{a+(b+c)}
=
(a+b+c)(ab+be+ca)
a
1
b+c x
bc
bc-
b+c
(b+c)²