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