分解因式:a³(b-c)+b³(c-a)+c³(a-b)=

3个回答

  • 原式=a^3(b-c)+b^3c-ab^3+ac^3-bc^3

    =a^3(b-c)+(b^3c-bc^3)+(ac^3-ab^3)

    =a^3(b-c)+bc(b^2-c^2)+a(c^3-b^3)

    =a^3(b-c)+bc(b+c)(b-c)-a(b-c)(b^2+bc+c^2)

    =(b-c)[a^3+bc(b+c)-a(b^2+bc+c^2)]

    =(b-c)[(a^3-ab^2)+bc(b+c)-a(bc+c^2)]

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

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

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

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

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

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

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