Answer
$(a)$ $$z=k\frac{x^3}{y^2}$$
$(b)$ $$z=k\frac{27x^3}{4y^2}$$
Work Step by Step
$(a)$
Applying direct variation of $x^3$ and inverse variation of $y^2$ gives us the following:
$z=k\times x^3 \times \frac{1}{y^2}=k\frac{x^3}{y^2}$
$(b)$ If we triple $x$ and double $y$, we will get:
$z=k\frac{(3x)^3}{(2y)^2}=k\frac{27x^3}{4y^2}$