Tõ x3+y3+z3 =((x+y)+z)3 =(x+y)3+3(x+y)2z +3(x+y)z2+z3
=x3+3x2y+3xy2+y3+3(x+y)z(x+y+z)+z3
Mµ x+y+z =1, x3+y3+z3 =1 nªn
3x2y+3xy2+3(x+y)z = 0
3x(x+y) +3(x+y)z=0
3(x+y).(xy+z)=0
Suy ra x+y=0 x= -y z = 1 suy ra P=1
xy+z=0 suy ra xy+1-x-y =0(tõ x+y+z=1 ta suy ra z = 1-x-y) x(y-1) – (y-1) =0
(y-1).(x-1) = 0 suy ra x = 1 y+z = 0 P=1 Y=1 x+z = 0 P=1