Tõ x 3 +y 3 +z 3 =((x+y)+z) 3 =(x+y) 3 +3(x+y) 2 z +3(x+y)z 2 +z 3 =x 3 +3x 2 y+3xy 2 +y 3 +3(x+y)z(x+y+z)+z 3 Mµ x+y+z =1, x 3 +y 3 +z 3 =1 nªn 3x 2 y+3xy 2 +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