a^3+b^3-c^3+3abc=?
回答 (3)
2016-07-30 2:05 pm
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
Then, a³ + b³ - c³ + 3abc
= a³ + b³ + (-c)³ - 3ab(-c)
= [a + b + (-c)] [a² + b² + (-c)² - ab - b(-c) - (-c)a]
= (a + b - c)(a² + b² + c² - ab + bc + ca)
Then, a³ + b³ - c³ + 3abc
= a³ + b³ + (-c)³ - 3ab(-c)
= [a + b + (-c)] [a² + b² + (-c)² - ab - b(-c) - (-c)a]
= (a + b - c)(a² + b² + c² - ab + bc + ca)
2016-07-30 1:59 pm
If you put c=a+b and simplify the result is 0.
Therefore (a+b-c) is a factor.
Therefore (a+b-c) is a factor.
2016-07-30 1:47 pm
uhh. ok.
收錄日期: 2021-04-18 15:27:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160730054606AAVwtyZ