Question

因式分解：a^2(b-c)+b^2(b-c)+c^2(a-b) ：(ab+1)(a+1)(b+1)+ab

Answer 1

a²(b-c)+b²(c-a)+c²(a-b)
=a²b-a²c+b²c-ab²+c²a-c²b
=ab(a-b)-c(a+b)(a-b)+c²(a-b)
=(a-b)[ab-c(a+b)+c²]
=(a-b)[ab-ac-bc+c²]
=(a-b)[b(a-c)-c(a-c)]
=(a-b)(a-c)(b-c)

(a+1)(b+1)(ab+1)+ab
=(ab+a+b+1)(ab+1)+ab
=(ab+1)^2+(a+b)(ab+1)+ab
=(ab+1)^2+a(ab+1)+b(ab+1)+ab
=(ab+1)^2+a(ab+b+1)+b(ab+1)
=(ab+1)(ab+1+b)+a(ab+b+1)
=(ab+b+1)(ab+a+1)