Question

若x+y+z=6,xy+yz+zx=11,x^3+y^3+z^3-3xyz

Answer 1

(x+y+z)²=6²

x²+y²+z²+2(xy+yz+zx)=36

x²+y²+z²=36-22=14

x³+y³+z³-3xyz

=(x³+3x²y+3xy²+y³+z³)-(3xyz+3x²y+3xy²)

=[(x+y)³+z³]-3xy(x+y+z)

=(x+y+z)(x²+y²+2xy-xz-yz+z²)-3xy(x+y+z)

=(x+y+z)(x²+y²+z²+2xy-3xy-xz-yz)

=(x+y+z)(x²+y²+z²-xy-yz-xz)

=18