1、已知x^2+y^2+6x-4y+13=0,则(x+y)^2006的值为-----.

x^2+y^2+6x-4y+13=0
(x^2+6x+9)+(y^2-4y+4)=0
(x+3)^2+(y-2)^2=0
x+3=0 y-2=0
x=-3 y=2
(x+y)^2006=1
x^2-x+1=0
(x+1)(x^2-x+1)=0
x^3+1=0
x^3=-1
x=-1
x^2001=-1
x^2+y^2+z^2-xy-yz-zx
=(2x^2+2y^2+2z^2-2xy-2yz-2zx)/2
=[(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)]/2
=[(x-y)^2+(y-z)^2+(z-x)^2]/2
=[1+1+1]/2
=1.5
-4m^3+16m^2-12m
=-4m(m^2-4m+3)
=-4m(m-1)(m-3)