$y$ 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.

### What about inverse proportion?

Same deal, only you flip the second fraction upside down!

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!

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## Cav

Here’s what I think: I love the method for speed, but wouldn’t teach it this way until the class understood, as it could cause issues later in when they need to form differential equations. It is a method that could be taught, however, last minute to those who don’t understand to boost test scores. But that isn’t something I’m keen on, and us almost a “necessary evil” by product of the examination system.

## Colin

Although, if you do DE the Ninja’s way, it all drops out nicely đ He’s declared war on arbitrary constants.