Question

若x+y+z=2,x^2+y^2+z^2=2,1/x+1/y+1/z=1/3则 x^3+y^3+z^3=

Answer 1

若x+y+z=2,x^2+y^2+z^2=2,1/x+1/y+1/z=1/3,xy+yz+xz=1,xyz=3则 x^3+y^3+z^3=(x^3+y^3+z^3-3xyz)+3xyz=(x+y+z)[(x^2+y^2+z^2)-(xy+yz+xz)]+3xyz=2*(2-1)+3*3=11