Question

(x^2)+(y^2)+(z^2)^-2(x+y+z)+3=0 求x^3+y^3+z^3+3xyz=

Answer 1

(x^2)+(y^2)+(z^2)-2(x+y+z)+3=0 求x^3+y^3+z^3+3xyz=

(x^2)+(y^2)+(z^2)-2(x+y+z)+3=0

(x^2-2x+1)+(y^2-2y+1)+(z^2-2z+1)=0

(x-1)^2+(y-1)^2+(z-1)^2=0

(x-1)^2=0 (y-1)^2=0 (z-1)^2=0

x=1 y=1 z=1

x^3+y^3+z^3+3xyz

=1+1+1+3=6