(1)已知a+b+c=6,a2+b2+c2=14,a3+b3+c3=36,求abc的值

4个回答

  • (1) (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=36

    ab+ac+bc=11

    (a+b+c)^3=a^3+b^3+c^3+6abc+3ab^2+3a^2b+3a^2c+3ac^2+3bc^2+3b^2c

    =14+6abc+18(a^2+b^2+c^2)-3(a^3+b^3+c^3)=14+6abc+18*14-3*36=216

    abc=29/3

    (2)(a-2b+c)(a+2b-c)-(a+2b+c)2

    =a^2-(2b-c)^2-(a^2+4b^2+c^2+4ab+2ac+4bc)

    =a^2-4b^2+4bc-c^2-(a^2+4b^2+c^2+4ab+2ac+4bc)

    =-8b^2-2c^2-4ab-2ac

    (3)(x+y)4(x-y)4=(x^2-y^2)^4

    (4)(a+b+c)(a2+b2+c2-ab-ac-bc)

    =a^3+b^3+c^3-3abc

    (5)(x+y+z)(x-y+z)(-x+y+z)(x+y-z)

    =[(x+z)^2-y^2][y^2-(x-z)^2]

    =[x^2+2xz+z^2-z^2+x^2][z^2-x^2-x^2-z^2+2xz]

    =(2x^2+2xz)(-2x^2+2xz)

    =2x(x+z)*(-2x)(x-z)

    =-4x^2(x^2-z^2)

    =4x^2y^2