Ta có \(x^3+y^3+z^3=3xyz\)

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

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

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

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]=0\)(Nhân hai vế với 2)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

Tới đây bạn xét hai trường hợp nhé :)

(x+y+z)((X+Y)^2-Z(X+Y))-3XY(X+Y+Z)

=(X+Y+Z)(X^2+2XY+Y^2-XZ-YZ-3XY)

=(X+Y+Z)(X^2+Y^2+Z^2-XZ-YZ-XY)

Ta có : \(x+y+z=0\Rightarrow-x-y=z\)

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

\(=-3x^2y-3xy^2=3xy\left(-x-y\right)=3xyz\) (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 
ok,xong r

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

\(\Rightarrow x^3+y^3+z^3-3xyz=0\)

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

\(\Rightarrow\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)=0\)

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

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

\(\Rightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\) (do \(x+y+z\ne0\))

\(\Rightarrow\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}\)\(\Rightarrow\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\)\(\Rightarrow\begin{cases}x=y\\y=z\\z=x\end{cases}\)\(\Rightarrow x=y=z\)

\(\Rightarrow P=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{1}\right)\left(1+\frac{1}{1}\right)=2\cdot2\cdot2=8\)

8

ta có x³+y³+z³=(x+y+z)(x²+y²+z²-xy-yz-xz)+3xyz(hằng đẳng thức)

Theo đề bài ->x+y+z=0

xem lại đề, chỗ 3xy²

Ta có:\(x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)

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

\(\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2xz-2yz\right)=0\)

\(\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\right]=0\)

\(x+y+z=0\)hoặc \(x=y=z\)(Đpcm)

a, \(x+y+z=0\)

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

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

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

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

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

Cảm ơn ạ