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

(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