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

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

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

ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)

\(\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=0\)

=> x + y + z = 0

Lai co: x³ + y³ +z³ - 3xyz = (x+y+z).(x²+y²+z² - xy - yz - zx)

             x³ + y³ + z³ - 3xyz = 0

=> x³ + y³ + z³ = 3xyz

ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)

=> 1/xy + 1/yz + 1/xz = 0

=> x + y + z = 0

Lai co: x³ + y³ +z³ - 3xyz = (x+y+z).(x²+y²+z² - xy - yz - zx)

             x³ + y³ + z³ - 3xyz = 0

=> x³ + y³ + z³ = 3xyz

Ta có : x+y+z = 0

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

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

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

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

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

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

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

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 )

x³ + y³ + z³ = 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

Lời giải :

\(x+y+z=0\)

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

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

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

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

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

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