Toán Lớp 8: Chứng minh: $x^{3}$ + $y^{3}$ + $z^{3}$ = $3xyz$ thì $a=b=c$ ; $a + b+ c = 0$

Question

Toán Lớp 8:  Chứng minh: $x^{3}$ + $y^{3}$ + $z^{3}$ = $3xyz$   thì   $a=b=c$  ;  $a + b+ c = 0$

ngoclanhoa · Answer

$  $   x^3 + y^3 + z^3 = 3xyz   -> x^3 + y^3 +z^3 - 3xyz=0   -> (x+y)^3 - 3xy (x+y) + z^3 - 3xyz=0   -> (x+y+z)[(x+y)^2 - (x+y)z + z^2] - 3xy (x+y+z)=0   -> (x+y+z) (x^2 + 2xy + y^2 - xz - yz +z^2) - 3xy (x+y+z)=0   -> (x+y+z) (x^2 + y^2 +z^2 +2xy - xz - yz - 3xy)=0   -> (x+y+z)(x^2 + y^2+z^2 - xy - yz - xz)=0   -> 1/2 (x+y+z) (2x^2 +2y^2 +2z^2 - 2xy - 2yz - 2xz)=0   -> 1/2 (x+y+z)[(x^2 - 2xy + y^2)+(y^2 -2yz +z^2) + (x^2 - 2xz +z^2)]=0   -> 1/2 (x+y+z) [(x-y)^2 + (y-z)^2 + (x-z)^2]=0   ->  ( left[  begin{array}{l}x+y+z=0 (1)  (x-y)^2 +(y-z)^2 + (x-z)^2=0 end{array}  right. )    $ bullet$ (x-y)^2 + (y-z)^2 + (x-z)^2=0   Với mọi x,y,z c&oacute; : $ begin{cases} (x-y)^2&ge;0  (y-z)^2 &ge;0  (x-z)^2&ge;0 end{cases}$   -> (x-y)^2 + (y-z)^2 + (x-z)^2 &ge;0&forall;x,y,z    Dấu '=' xảy ra khi :   &harr; $ begin{cases} (x-y)^2=0  (y-z)^2=0  (x-z)^2=0  end{cases}$   &harr; $ begin{cases} x-y=0  y-z=0  x-z=0  end{cases}$   &harr; $ begin{cases} x=y  y=z  x=z  end{cases}$   &harr; x=y=z (2)   Từ (1), (2)   -> x=y=z, x+y+z=0