Question

Phân tích a²b²(a-b)+b²c²(b-c)+c²a²(c-a) thành nhân tử

Answer 1

a\(^2\)b\(^2\)(a−b)+b\(^2\)c\(^2\)(b−c)+c\(^2\)a\(^2\)(c−a)

=a\(^2\)b\(^2\)(a−b)−b\(^2\)c\(^2\)(a−b+c−a)+c\(^2\)a\(^2\)(c−a)

=a\(^2\)b\(^2\)(a−b)−b\(^2\)c\(^2\)(a−b)−b\(^2\)c\(^2\)(c−a)+c\(^2\)a\(^2\)(c−a)

=b\(^2\)(a−b)(a\(^2\)−c\(^2\))−c\(^2\)(c−a)(b\(^2\)−a\(^2\))

=b\(^2\)(a−b)(a−c)(a+c)−c\(^2\)(c−a)(b−a)(b+a)

=(a−b)(a−c)(ab\(^2\)+b\(^2\)c)−(a−c)(a−b)(ac\(^2\)+bc\(^2\))

=(a−b)(a−c)(ab\(^2\)+b\(^2\)c−ac\(^2\)−bc\(^2\))

=(a−b)(a−c)[(ab\(^2\)−ac\(^2\))+bc(b−c)]

=(a−b)(a−c)[(ab+ac)(b−c)+bc(b−c)]

=(a−b)(a−c)(b−c)(ab+ac+bc)