Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)

nên \(\dfrac{x}{10}=\dfrac{y}{15}\)(1)

Ta có: \(\dfrac{y}{5}=\dfrac{z}{4}\)

nên \(\dfrac{y}{15}=\dfrac{z}{12}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

Đặt \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=10k\\y=15k\\z=12k\end{matrix}\right.\)

Ta có: xyz=1800

\(\Leftrightarrow1800k^3=1800\)

\(\Leftrightarrow k^3=1\)

\(\Leftrightarrow k=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=10\cdot1=10\\y=15\cdot1=15\\z=12\cdot1=12\end{matrix}\right.\)

\(x,y,z>0\)

Áp dụng BĐT Caushy cho 3 số ta có:

\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)

\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)

\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)

Áp dụng BĐT Caushy-Schwarz ta có:

\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)

\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)

\(P=0\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=0\)