jmc

Taking the product of the equations gives $$\begin{array}{rl}xy\cdot xz\cdot yz& =20\sqrt[3]{2}\cdot 35\sqrt[3]{2}\cdot 14\sqrt[3]{2}\\ (xyz{)}^2& =2^4\cdot 5^2\cdot 7^2\\ xyz& =2^2\cdot 5\cdot 7=\pm 140\end{array}$$ Since we were given that $x$, $y$, and $z$ are positive, we can conclude that $xyz=\boxed{140}$.