Lời giải:
Ta thấy, với mọi $x,y,z$ là số thực thì:

$(x-y+z)^2\geq 0$

$\sqrt{y^4}\geq 0$

$|1-z^3|\geq 0$

$\Rightarrow (x-y+z)^2+\sqrt{y^4}+|1-z^3|\geq 0$ với mọi $x,y,z$

Kết hợp $(x-y+z)^2+\sqrt{y^4}+|1-z^3|\leq 0$

$\Rightarrow (x-y+z)^2+\sqrt{y^4}+|1-z^3|=0$

Điều này xảy ra khi: $x-y+z=y^4=1-z^3=0$

$\Leftrightarrow y=0; z=1; x=-1$

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

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mn giúp mình với

xét hiệu x³+y³+z³-3xyz

=(x+y)³+z³-3xy(x+y)-3xyz

=(x+y+z)³-3(x+y+z)(x+y)z-3xy(x+y+z)

=0       vì x+y+z=0

=>x³+y³+z³=3xyz

=>đpcm

\(x+y+z=0\)

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

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

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

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

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

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

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)

Áp dụng bđt AM-GM:

\(x+y\ge2\sqrt{xy}\)

\(y+z\ge2\sqrt{yz}\)

\(z+x\ge2\sqrt{xz}\)

Nhân theo vế:\(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8xyz\)

\("="\) khi x=y=z

Khi đó hiển nhiên \(x^3+y^3+z^3=3xyz\)

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

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

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

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

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

=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