Do x\(^3\)+y\(^3\)+z\(^3\)=3xyz\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+y+z=0\\x=y=z\end{array}\right.\)

TH1:x+y+z=0\(\Rightarrow P=\frac{xyz}{\left(-z\right)\left(-y\right)\left(-x\right)}=-1\)

TH2:x=y=z\(\Rightarrow P=\frac{xyz}{8xyz}=\frac{1}{8}\)

Ta có: x³ + y³ + z³ = 3xyz

x³ + y³ + z³ - 3xyz = 0

x³ + 3x²y + 3xy² + y³ + z³ - 3xy(x + y) - 3xyz = 0

(x + y)³ + z² - 3xy(x + y + z) = 0

(x + y + z)[(x + y)² - (x + y)z + z²] - 3xy(x + y + z) = 0

(x + y + z)(x² + 2xy + y² - xz - yz + z²) - 3xy(x + y + z) = 0

(x + y + z)(x² + 2xy + y² - xz - yz + z² - 3xy) = 0

(x + y + z)(x² + y² + z² - xz - yz - xy) = 0

=> x + y + z = 0 hoặc x² + y² + z² - xz - yz - xy = 0

+) Với x + y + z = 0 

<=> x + y = -z, x + z = -y, y + z = -x

Thay x + y = -z, x + z = -y, y + z = -x vào P, ta có:

\(P=\frac{xyz}{\left(-z\right)\left(-x\right)\left(-y\right)}=-1\)

+) Với x² + y² + z² - xz - yz - xy = 0

=> 2x² + 2y² + 2z² - 2xz - 2yz - 2xy = 0

=> (x² - 2xy + y²) + (x² - 2xz + z²) + (y² - 2yz + z²) = 0

=> (x - y)² + (x - z)² + (y - z)² = 0

=> (x - y)² = 0 và (x - z)² = 0 và (y - z)² = 0

=> x = y và x = z và y = z

=> x = y = z

Thay x = y = z vào P, ta có:

\(P=\frac{xxx}{\left(x+x\right)\left(x+x\right)\left(x+x\right)}=\frac{x^3}{\left(2x\right)^3}=\frac{x^3}{8x^3}=\frac{1}{8}\)

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

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

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

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

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

\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)

\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)