$y$ is directly proportional to $x^3$, you say? And when $x = 4$, $y = 72$? Well, then.

The traditional method is to say:

$y = kx^3$ and substitute in what you know.

$72 = 64k$ $k = \frac{72}{64} = \frac{9}{8}$

That gives $y = \frac98 x^3$. Easy enough.

### But you can do it without the $k$

Rather than introduce an arbitrary constant, you can do this directly by writing it as:

$\frac {y}{y_0} = \frac {x^3}{x_0^3}$, where $x_0$ and $y_0$ are the numbers you’re given:

$\frac y {72} = \frac{x^3}{64}$

$y = \frac{72}{64}x^3$, which cancels down to the same thing as before.

If you know $B$ is inversely proportional to $r^2$, and when $r = 3,$ $B = 1$, you can do this:
$\frac {B}{B_0} = \frac{r_0^2}{r^2}$ $\frac{B}{1} = \frac{9}{r^2}$ $B = \frac{9}{r^2}$
I find that a bit easier to get my head around than setting up an arbitrary $k$ and finding its value - but I’d be interested to hear what you think!