$\text{1. x³+y³+z³=3xyz}$
$\text{⇔x³+y³+z³-3xyz=0}$
$\text{⇔(x+y)³+z³-3xy(x+y)-3xyz=0}$
$\text{⇔(x+y+z).[(x+y)²-(x+y).z+z²)]-3xy(x+y+z)=0}$
$\text{⇔(x+y+z).(x²+y²+z²-xy-yz-zx)=0}$
$\text{⇔(x+y+z).(2.x²+2y²+2.z²-2xy-2yz-2zx)=0}$
$\text{⇔(x+y+z).[(x-y)²+(y-z)²+(z-x)²]=0}$
$\text{VÌ x,y,z đội 1 khác nhau nên}$
$\text{(x-y)²+(y-z)²+(z-x)² khác 0}$
$\text{Suy ra x+y+z=0}$
$\text{2.B=4.(x³+y³)-6.(x²+y²)}$
$\text{⇔ B=4.(x+y)(x²-xy+y²)-6.(x²+y²)}$
$\text{⇔B=4.(x²-xy+y²)-6.(x²+y²) (do x+y=1)}$
$\text{⇔B=-2.x²-4xy-2y²}$
$\text{⇔B=-2(x+y)²=-2}$
