Question

一道因式分解题：a^2[b-c]+b^2[c-a]+c^2[a-b] *注：a的n次方表示为"a^b"

Answer 1

原式=ba^2-ca^2+cb^2-ab^2+c^2[a-b]

=(ba^2-ab^2)-(ca^2-cb^2)+c^2[a-b]

=ab(a-b)-c(a^2-b^2)+c^2[a-b]

=ab(a-b)-c(a-b)(a+b)+c^2[a-b]

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

=(a-b)[(ab-ac)-(bc-c^2)]

=(a-b)[a(b-c)-c(b-c)]

=(a-b)[(a-c)(b-c)]

=(a-b)(a-c)(b-c)