Why does $(3\sqrt3)^2 = 27$? How does $(3\sqrt3)^2 = 27$?
I've tried to solve this using binomial expansion and using the FOIL method from which I obtain $9 + 3\sqrt3 +3\sqrt3 + 3$.
it has been a while since I've done this kind of thing so it may be something obvious that I can't see.

Other people have basically already said it, but the FOIL is only something that you can apply when you have expressions of exactly this form:
$$
(a + b)(c+d).
$$
In this case you do: (F)irst (O)utside (I)nside (L)ast: The First part of $(a + b)^2$ is $ac$. The Outside is $ad$, the Inside if $bc$ and the Last is $bd$. In all you get
$$
ac + ad + bc + bd.
$$
Now consider what happens when you have
$$
(a+b)^2 = (a+b)(a+b).
$$
In this case you get
$$
aa+ ab + ba + bb = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2.
$$
Now, you are asking about what to do with
$$
(3\sqrt{3})^2
$$
Note that this is not of exactly the form $(a+b)(c+d)$ or $(a+b)^2$. You have the product of $3$ and $\sqrt{3}$ and to use FOIL you would need the sum $3+\sqrt{3}$. Therefore it doesn't make sense to use FOIL here. It simply isn't applicable (unless you rewrite the expression first as explain in one of the other answers).

So what do you do in this case? Your expression is exactly of the form
$$
(ab)^2
$$
with $a=3$ and $b = \sqrt{3}$. And then we have a rule that says that
$$
(ab)^2 = a^2b^2.
$$
So you get
$$
(3\sqrt{3})^2 = 3^2(\sqrt{3})^2 = 9\cdot 3 = 27.
$$
And that is it.

You could also do it slightly differently by using the rule/definition
$$
(ab)^2 = abab 
$$
So
$$
(3\sqrt{3})^2 = 3\sqrt{3}\cdot3\sqrt{3} = 3\cdot 3\cdot \sqrt{3}\sqrt{3} = 9 \sqrt{3\cdot 3} = 9\sqrt{9} = 9\cdot 3 = 27.
$$
Here we also used that the rule $\sqrt{c}\sqrt{d} = \sqrt{cd}$.
