Ta có x³ + y³ + z³ - 3xyz=\(\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(\left(x+y\right)^2-z\left(x+y\right)+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz-3xy\right)=0\)

Vì x³ + y³ + z³ - 3xyz=0 nên x³ + y³ + z³=3xyz

x+y+z= 0
x+y=-z
(x+y)^3 =-z^3
x^3 +y^3 +3xy(x+y) =-z^3
x^3 +y^3 +3xy(-z) =-z^3
x^3 +y^3 -3xyz =-z^3
x^3 +y^3 + z^3 =3xyz => dpcm

Ta có: \(x+y+z=0\Leftrightarrow x+y=-z\)

\(\Leftrightarrow\left(x+y\right)^3=\left(-z\right)^3\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=-z^3\)

\(\Leftrightarrow x^3+y^3+z^3=-3x^2y-3xy^2\)

\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(x+y\right)\)

\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(-z\right)=3xyz\)(đpcm)

ta có 

x+y + z = 0

=> x+y = -z

=> (x+y) ^3 = (-z)^3 

=> x^3 + y^3 + 3xy(x+y) = -z^3 

=> x^3 + y^3 + z^3 = -3xy(x+y)

=> x^3 + y^3 + z^3 = 3xyz ( đpcm)

= ( x³ + 3x² y +3xy² + y³ ) + z³ - 3 x ² y - 3xy² - 3xyz

= ( x + y ) ³ + z³ ] - 3xy x ( x + y + z )

= ( x + y + z ) x [ ( x + y ) ² - z ( x + y ) + z² ] - 3xy x ( x + y + z )

= ( x + y + z ) x ( x² + 2xy + y² + zx - zy + z² - 3xy )

= ( x + y + z ) . ( x² + y² + z² - xy - yz - zx )

Liên quan thế từ x + y + z sang a +b +c

Xét \(A=x^3+y^3+z^3-3xyz\)

\(=\left(x^3+3x^2y+3xy^2+y^3\right)+z^3-3x^2y-3xy^2-3xyz\)

\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz+yz+z^2\right)-3xy\left(x+y+z\right)\)

Với \(x+y+z=0\) thì  \(A=0.\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy.0=0\)

\(A=x^3+y^3+z^3-3xyz=0\Rightarrow x^3+y^3+z^3=3xyz\) (đpcm)

x+y+z=0=> (x+y+z)(x²+y²+z²-xy-yz-zx)=0 (*)

Nhân (*) ra được :

x³+y³+z³-3xyz=0<=> x³+y³+z³= 3xyz(đpcm)

Ta có : x + y + z = 0 => x + y = -z => (x + y)³ = (-z)³

=> x³ + 3x²y + 3xy² + y³ = (-z)³

=> x³ + y³ + z³ + 3x²y + 3xy² = 0

=> x³ + y³ + z³ + 3xy(x + y) = 0

Mà x + y = -z

Nên :  x³ + y³ + z³ + 3xy(-z) = 0

=>  x³ + y³ + z³ - 3xyz = 0

=> A = 0

Vậy x + y + z = 0 thì A = 0 (đpcm)

Từ:

 x + y + z = 0 

=> x + y = -z 
<=> (x + y)^3 = (-z)^3 
<=> x^3 + 3x^2y + 3xy^2 + y^3 = -z^3 
<=> x^3 + y^3 + z^3 = -3x^2y - 3xy^2 
<=> x^3 + y^3 + z^3 = -3xy(x+y) 
<=> x^3 + y^3 + z^3 = -3xy(-z) 
<=> x^3 + y^3 + z^3 = 3xyz 

P/s: Tham khảo nha

\(x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(-z\right)\)

\(\Leftrightarrow x^3+y^3+z^3=-3\left(x+y\right)\)

\(\Leftrightarrow x^3+y^3+z^3=-3x^2y-3xy^2\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=-z^3\)

\(\Leftrightarrow\left(x+y\right)^3=\left(-z\right)^3\)

\(\Leftrightarrow x+y=-z\)

\(\Leftrightarrow x+y+z=0\)