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

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

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

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

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

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

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

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

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

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

Đặt : A = x³ + y³ - z³

Ta có : A= x³ + y³ - z³

A= ( x + y)³ - 3xy(x + y) - z³

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

Thay x + y = z vào A ta có :

A = ( z - z).( z² + z.z + z² ) - 3xyz

A = 0.( z² + z.z + z² ) - 3xyz

A= -3xyz ( đpcm )

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

\(b,x^3+y^3+z^3-3xyz\\ =\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz+2xy-3xy\right)\\ =0\left(x^2+y^2+z^2-xz-yz-xy\right)=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)

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)

x+y+z=0

<=>x+y=-z

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

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

Mà x+y=-z

=>đccm