1;\(A=x^3+y^3+z^3-3xyz\)

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

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

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

2;Nếu A = 0

Điều ngược lại đúng khi x^2+y^2+z^2-xy-yz-xz khác 0

Ta đi chứng minh A phụ thuộc vào x+y+z

\(A=x^3+y^3+z^3-3xyz.\)

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

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

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

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

Mà x^2+y^2+z^2-xy-yz-xz>0

nên  x+y+z =0 thì A=0

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

\(xet:x=y=z=1\Rightarrow A=1+1+1-3=0\Rightarrow dieunguoclaichuachacdadung\)

nếu x+y+z=0 thì x^3+y^3+z^3=3xyz

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

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

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

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

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

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

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

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

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

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

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

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)