Chapter 6 - Section 6.7 - Variation and Problem Solving - Exercise Set - Page 398: 37

$k=3$

Since $y$ varies jointly as $x$ and the cube of $z$, then, $ y=kxz^3 .$ Substituting the given values of the variables, then the value of $k$ is, \begin{array}{l} 120=k(5)(2)^3 \\\\ 120=k(5)(8) \\\\ 120=k(40) \\\\ \dfrac{120}{40}=k \\\\ 3=k \\\\ k=3 .\end{array}