(3)
(a+b)(b+c)(c+a)+abc
=(a+b){bc+ab+c²+ca)+abc
=(a+b)(ab+bc+ca)+(a+b)c²+abc
=(a+b)(ab+bc+ca)+c²a+bc²+abc
=(a+b)(ab+bc+ca)+c(ab+bc+ca)
=(a+b+c)(ab+bc+ca)
Mathematics
Junior High
3番をお願いします…
しコココン
31 次の式を因数分解せよ。
(1) ce(⑫ー/)十cg(c一の)斑25(2ーめの
(⑫②) 2が(2十の十7c(5十c)十cg(c填の)十2g0c
(3 (Z十の(6十の)(c十の十g0c
Answers
Were you able to resolve your confusion?
Users viewing this question
are also looking at these questions 😉