Answer
$(a)$ $$z=kx^3y^5$$
$(b)$ $$864$$
Work Step by Step
$(a)$
According to the problem, $z$ is jointly proportional to $x^3$ and $y^5$, so we can write:
$z=kx^3y^5$
$(b)$ If we triple $x$ and double $y$, we will get:
$z=k(3x)^3(2y)^5=k\times27x^3\times 32y^5=864\times kx^3y^5$
It changes by a factor of $864$