BÀI TOÁN CẦN NHỚ
I/Phân tích đa thức sau thành nhân tử :
x3+y3+z3 – 3xyz
Ta có : x3+y3+z3 – 3xyz
= (x + y )3 + z3 – 3xyz
@(x + y )3 = x3 + 3x2y + 3xy2 + y3
– 3x2y – 3xy2
=( x+y+z )(x2+y2+z2+2xy-xz-yz) – 3xy(x+y+z)
@ (x + y )3 + z3
=( x+y+z )[ (x+y)2 – (x+y)z +z2 ]
=( x+y+z )(x2+y2+z2+2xy-xz-yz)
=( x + y + z )( x2 + y2 + z2 – xy – xz – yz )
@( x2 + y2 + z2 – xy – xz – yz )
=( x+y+z )(x2 +2xy +y2-xz-yz+z2)
II/ ÁP DỤNG :
1/ Tìm : x ; y ; z để x3+y3+z3 = 3xyz ( hay x3+y3+z3 – 3xyz = 0 )
Ta có : x3+y3+z3 = 3xyz
Hay : (x + y )3 + z3 – 3xyz = 0
( x+y+z )(x2+y2+z2+2xy-xz-yz) – 3xy(x+y+z) = 0
( x + y + z )( x2 + y2 + z2 – xy – xz – yz ) = 0
@Vậy : x + y + z = 0 hoặc x = y = z thì : x3 + y3 + z3 = 3xyz ( hay x3 + y3 + z3 – 3xyz = 0 )
2/ CMR: Nếu x = y = z thì :
x3+y3+z3 = 3xyz ( hay x3+y3+z3 – 3xyz = 0 )
3/ CMR: Nếu x + y + z = 0 thì :
x3+y3+z3 = 3xyz ( hay x3+y3+z3 – 3xyz = 0 )
=( x+y+z )(x2+y2+z2+2xy-xz-yz) – 3xy(x+y+z)
=( x + y + z )( x2 + y2 + z2 – xy – xz – yz )
Vậy : x3+y3+z3 = 3xyz ( hay x3+y3+z3 – 3xyz = 0 )
= 0 ( Vì x – y = 0 ; y – z = 0 ; z - x = 0 )

Xét : x3+y3+z3 – 3xyz
= 0 ( Vì x + y + z = 0 )
Vậy : x3+y3+z3 = 3xyz ( hay x3+y3+z3 – 3xyz = 0 )
=( x + y + z )( x2 + y2 + z2 – xy – xz – yz )