因式分解 (ab+cd)(a²-b²+c²-d²)+(ac+bd)(a&sup2

1个回答

  • 1)原式

    =(a^3b+a^3c)+(a^2bd+a^2cd)+(abc^2+ab^2c-ab^3-ac^3-abd^2-acd^2)-(bd^3+cd^3)-(b^2cd+bc^2d)+(b^3d+c^3d)

    =a^3(b+c)+a^2d(b+c)+a((bc^2+b^2c)-(b^3+c^3)-(bd^2+cd^2))-d^3(b+c)-bcd(b+c)+d(b^3+c^3)

    =a^3(b+c)+a^2d(b+c)+a((bc^2+b^2c)-(b^3+c^3)-(bd^2+cd^2))-d^3(b+c)-bcd(b+c)+d(b+c)(b^2-bc+c^2)

    =(b+c)((a^3-d^3)+(a^2d-ad^2)+(2abc-2bcd)-(ab^2-b^2d)-(ac^2-c^2d))

    =(b+c)((a-d)(a^2+ad+d^2)+ad(a-d)+2bc(a-d)-b^2(a-d)-c^2(a-d))

    =(b+c)(a-d)((a^2+2ad+d^2)-(b^2-2bc+c^2))

    =(a-d)(b+c)((a+d)^2-(b-c)^2)

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

    2)原式

    =ax^3+x^3-ax^2y+bx^2y+axy^2-bxy^2+by^3+y^3+(ay^3+bx^3-ay^3-bx^3)

    =(ax^3+ay^3)+(bx^3+by^3)+(x^3+y^3)-((ax^2y-axy^2+ay^3)+(bx^3-bx^2y+bxy^2))

    =a(x^3+y^3)+b(x^3+y^3)+(x^3+y^3)-(ay(x^2-xy+y^2)+bx(x^2-xy+y^2))

    =(a+b+1)(x+y)(x^2-xy+y^2)-(ay+bx)(x^2-xy+y^2)

    =((a+b+1)(x+y)-(ay+bx))(x^2-xy+y^2)

    =(ax+x+by+y)(x^2-xy+y^2)

    3)原式

    =(x^4yz-y^3z^3)-(x^3y^3+x^3z^3)+(xy^4z+xyz^4)

    =yz(x^4-y^2z^2)-x^3(y^3+z^3)+xyz(y^3+z^3)

    =yz(x^2+yz)(x^2-yz)-x(x^2-yz)(y^3+z^3)

    =(x^2-yz)(yz(x^2+yz)-x(y^3+z^3))

    =(x^2-yz)((x^2yz-xy^3)+(y^2z^2-xz^3))

    =(x^2-yz)(xy(xz-y^2)-z^2(xz-y^2))

    =(x^-yz)(y^2-zx)(z^2-xy)