若x+y+z=6,xy+yz+zx=11,x^3+y^3+z^3-3xyz

(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

x^3+y^3+z^3-3xyz
=[( x+y)^3-3x^2y-3xy^2]+z^3-3xyz
=[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)
=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)
=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2)-3xy(x+y+z)
=(x+y+z)(x...

由已知得 x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=36-22=14 ，
因此 x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=6*(14-11)=18 。