\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)-\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)+z^3\)
\(=\left(\frac{1}{x}+\frac{1}{y}\right)^3+\frac{1}{z^3}-\frac{3}{xy}\left(\frac{-1}{z}\right)\) (do \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\))
\(=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left[\left(\frac{1}{x}+\frac{1}{y}\right)^2-\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{z}+\frac{1}{z^2}\right]+\frac{3}{xyz}\)
\(=\frac{3}{xyz}\)
\(\Rightarrow P=\frac{2017}{3}.xyz.\frac{3}{xyz}=2017\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{1}{x}=-\left(\frac{1}{y}+\frac{1}{z}\right).P=\frac{2017}{3}xyz\left[-\left(\frac{1}{y}+\frac{1}{z}\right)^3+\frac{1}{y^3}+\frac{1}{z^3}\right]=-\frac{2017}{3}xyz\left(\frac{3}{yz^2}+\frac{3}{zy^2}\right)=-2017xyz\left(\frac{z+y}{z^2y^2}\right)=-2017\left(\frac{xyz^2+xy^2z}{y^2z^2}\right)=-2017\left(\frac{x}{y}+\frac{x}{z}\right)=-2017x\left(\frac{1}{y}+\frac{1}{z}\right)=-2017.\left(-\frac{1}{x}\right)x=2017\)