Ta có: \(\frac{x^3+y^3+z^3-3xyz}{x+y+z}\)

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

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

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

\(=x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\left(\forall x,y,z\right)\)

=> đpcm

lllllllllllllllllllllllllllllllllllllllllllllllllllllll

mn giúp mình với

\(x+y+z=0\)

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

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

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

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

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

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

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

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

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

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

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

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

Có : x + y + z = 0

--> x + y = -z

--> (x+y)³ = (-z)³

--> x³ + y³ + 3xy(x+y) = (-z)³

--> x³ + y³  +3xy(-z) = (-z)³

--> x³ + y³ - 3xyz = (-z)³

--> x³ + y³ + z³ = 3xyz (đpcm)

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

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

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

=0

\(x+y+z=0\\ \Rightarrow x+y=-z\\ \Rightarrow\left(x+y\right)^3=\left(-z\right)^3\\ \Rightarrow x^3+3x^2y+3xy^2+y^3\\ \Rightarrow x^2+y^2+z^2=-3x^2y-3xy^2\\ \Rightarrow x^2+y^2+z^2=-3xy\left(x+y\right)\\ \Rightarrow x^2+y^2+z^2=-3xy\left(-z\right)=3xyz\\ \left(đpcm\right)\)

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)=0

x^3+y^3+z^3-3xyz=0

=> x^3+y^3+z^3=3xyz

kagamine rin len đúng rồi đó

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

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

Ta có: x+y+z=0⇔x+y=−z

⇔(x+y)³=(−z)³

⇔x³+3x²y+3xy²+y³=−z³

⇔x³+y³+z³=−3x²y−3xy²

⇔x³+y³+z³=−3xy(x+y)

⇔x³+y³+z³=−3xy(−z)=3xyz(đpcm)

ta có thể cm x^3+y^3+z^3=3xyz =>(x+y+z)(a^2+b^2+c^2-ab-ac-bc)=0

=>a^2+b^2+c^2-ab-ac-bc=0

nhân cả 2 vế với 2 ta đc

2.(x^2+y^2+z^2-xz-yz-yx)=2.0=0

=2x^2+2y^2+2z^2-2xy-2xz-2yz

=>(y^2-2yx+x^2)+(y^2-2xz+z^2)+(x^2-2xz+z^2)=0

<=> (y-x)^2+(y-z)^2+(x-z)^2=0

mà ta lại có  (y-x)^2>=0 ;  (y-z)^2>=0 ;  (x-z)^2>=0

 và (y-x)^2+(y-x)^2+(x-z)^2=0

 <=>(y-x)^2=0<=>y=x

  <=>(y-z)^2=0 <=>y=z

  <=>(x-z)^2=0<=>x=z

=>x=y=z